Sridharacharya Formula: Derivation & Examples

#Algebra
TL;DR
The Sridharacharya formula is the rule that solves any quadratic equation ax2+bx+c=0ax^2 + bx + c = 0, and it reads x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. It is the same result the rest of the world calls the quadratic formula, named for the Indian mathematician Sridharacharya, who gave the completing-the-square method behind it. The part under the square root, b2−4acb^2 - 4ac, is the discriminant, and its sign tells you how many real roots the equation has before you solve.
BT
Bhanzu TeamLast updated on September 10, 202610 min read

What Is The Sridharacharya Formula?

The Sridharacharya formula gives the exact solutions of any quadratic equation written in standard form, ax2+bx+c=0ax^2 + bx + c = 0, as long as a≠0a \neq 0. Known worldwide as the quadratic formula, it is:

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

To use it, read off the three numbers aa, bb, and cc from the equation, substitute them, and simplify. The ±\pm sign is doing real work: it splits into two calculations, one with a plus and one with a minus, which is why a quadratic usually has two solutions.

Table: The parts of the Sridharacharya formula, and what each symbol means.

Symbol

Meaning

aa

Coefficient of x2x^2, the quadratic term. Must be non-zero.

bb

Coefficient of xx, the linear term.

cc

The constant term (no xx attached).

xx

The unknown you are solving for; its values are the roots.

±\pm

Splits the formula into two roots, one using ++ and one using −-.

b2−4acb^2 - 4ac

The discriminant, DD. Its sign fixes how many real roots exist.

The roots are the values of xx that make the equation true. Graphically, they are the points where the curve y=ax2+bx+cy = ax^2 + bx + c meets the x-axis.

Where Does The Sridharacharya Formula Come From?

The formula is completing the square carried out once, in general, with the letters left in place of numbers. Sridharacharya's own method avoids fractions by multiplying through by 4a4a first, and that is the route shown here.

Start from the standard form.

ax2+bx+c=0ax^2 + bx + c = 0

Multiply every term by 4a4a. This clears the way to a clean square later.

4a2x2+4abx+4ac=04a^2x^2 + 4abx + 4ac = 0

Move the constant term to the right-hand side.

4a2x2+4abx=−4ac4a^2x^2 + 4abx = -4ac

Add b2b^2 to both sides. This is the step that turns the left side into a perfect square.

4a2x2+4abx+b2=b2−4ac4a^2x^2 + 4abx + b^2 = b^2 - 4ac

The left side now folds into a single squared bracket.

(2ax+b)2=b2−4ac(2ax + b)^2 = b^2 - 4ac

Take the square root of both sides. The ±\pm enters here, because a square root can be positive or negative.

2ax+b=±b2−4ac2ax + b = \pm\sqrt{b^2 - 4ac}

Subtract bb, then divide both sides by 2a2a.

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

That is the Sridharacharya formula, derived line by line from the equation it solves.

How Do You Use The Sridharacharya Formula?

To solve a quadratic, match it to ax2+bx+c=0ax^2 + bx + c = 0, list aa, bb, and cc with their signs, then substitute into the formula. The signs are where most marks are lost, so write them down before touching the formula.

Example 1: Solve 2x2−x−3=02x^2 - x - 3 = 0.

First identify the coefficients. Comparing with ax2+bx+c=0ax^2 + bx + c = 0 gives a=2a = 2, b=−1b = -1, and c=−3c = -3.

Compute the discriminant.

b2−4ac=(−1)2−4(2)(−3)=1+24=25b^2 - 4ac = (-1)^2 - 4(2)(-3) = 1 + 24 = 25

Since 25>025 > 0, there are two distinct real roots. Now substitute into the formula, taking care that −b=−(−1)=+1-b = -(-1) = +1.

x=−(−1)±252(2)=1±54x = \frac{-(-1) \pm \sqrt{25}}{2(2)} = \frac{1 \pm 5}{4}

Split the ±\pm into the two roots.

x=1+54=64=32,x=1−54=−44=−1x = \frac{1 + 5}{4} = \frac{6}{4} = \frac{3}{2}, \qquad x = \frac{1 - 5}{4} = \frac{-4}{4} = -1

Final answer: x=32x = \frac{3}{2} or x=−1x = -1. You can check either root by substituting it back: 2(32)2−32−3=92−32−3=02\left(\tfrac{3}{2}\right)^2 - \tfrac{3}{2} - 3 = \tfrac{9}{2} - \tfrac{3}{2} - 3 = 0.

What Does The Discriminant Tell You About The Roots?

The discriminant D=b2−4acD = b^2 - 4ac is the part of the formula under the square root, and its sign settles the nature of the roots before you finish solving. Reading it first tells you what kind of answer to expect.

Table: How the discriminant decides the nature of the roots.

Discriminant D=b2−4acD = b^2 - 4ac

Nature of the roots

D>0D > 0

Two distinct real roots

D=0D = 0

Two equal real roots (one repeated root)

D<0D < 0

No real roots (two complex conjugate roots)

When D=0D = 0, the square root vanishes and the formula collapses to x=−b2ax = \frac{-b}{2a}; the equation x2−6x+9=0x^2 - 6x + 9 = 0 is one such case, with x=3x = 3 counted twice. When D<0D < 0, the square root has no real value, so the roots of the quadratic equation are a pair of complex numbers rather than points on the x-axis.

Example 2: Solve x2−4x+1=0x^2 - 4x + 1 = 0.

Here a=1a = 1, b=−4b = -4, and c=1c = 1, so the discriminant is (−4)2−4(1)(1)=16−4=12(-4)^2 - 4(1)(1) = 16 - 4 = 12. That is positive, so expect two real roots, and because 1212 is not a perfect square, they will be irrational.

x=−(−4)±122(1)=4±122x = \frac{-(-4) \pm \sqrt{12}}{2(1)} = \frac{4 \pm \sqrt{12}}{2}

Simplify the surd: 12=23\sqrt{12} = 2\sqrt{3}.

x=4±232x = \frac{4 \pm 2\sqrt{3}}{2}

Divide the whole numerator, both terms, by 22.

x=2±3x = 2 \pm \sqrt{3}

Final answer: x=2+3x = 2 + \sqrt{3} or x=2−3x = 2 - \sqrt{3}.

Why Does The Sridharacharya Formula Always Work?

Factoring only lands when the roots happen to be tidy whole numbers or simple fractions. The Sridharacharya formula has no such limit, and there are three reasons worth naming.

  • It is completing the square, generalised. Every quadratic can be rewritten as a perfect square plus a constant, so every quadratic can be solved by the same square-root step. The formula just does that rearrangement once, with letters, so you never have to repeat it.

  • The ±\pm is the parabola's symmetry. A parabola is a mirror-symmetric curve, and its two roots sit at equal distances on either side of the line x=−b2ax = \frac{-b}{2a}. The plus and minus in the formula are that mirror, measured out by b2−4ac2a\frac{\sqrt{b^2 - 4ac}}{2a}.

  • The discriminant is the switch. Whether those two mirror points land on the x-axis, meet at it, or miss it entirely depends only on the sign of b2−4acb^2 - 4ac.

So the formula is not a lucky trick to memorise. It is what completing the square becomes when you carry it out for every quadratic at the same time.

Who Invented The Sridharacharya Formula?

The method behind the formula was written down in India more than a thousand years ago, long before algebra used the symbols we write today. The person it is named for very nearly slipped out of the record entirely.

Three Indian mathematicians shaped this corner of algebra:

  • Sridharacharya (c. 870–930 CE, India) wrote the practical mathematics texts Patiganita and Trisatika, and is remembered as one of the first to give a general rule for the quadratic.

  • Bhaskara II (1114–1185 CE, India), also called Bhaskaracharya, preserved Sridhara's rule and showed clearly that a single quadratic can have more than one solution.

  • Brahmagupta (598–668 CE, India) had given an even earlier rule for solving quadratics, written entirely in words, centuries before symbolic algebra existed.

Where Is The Sridharacharya Formula Used In The Real World?

Quadratics appear wherever a quantity rises and then turns back, and the formula is how the turning points get found.

  • Physics and engineering: the height of a thrown ball or a launched rocket is a quadratic in time, so projectile motion problems ask a quadratic question: at what times is the object at a given height?

  • Architecture and construction: the cable of a suspension bridge and the sweep of a parabolic arch follow quadratics, and finding where the curve meets a support means solving one.

  • Business and economics: revenue and profit models are often quadratic, and the break-even points, where profit equals zero, are exactly the roots.

  • Computer graphics and games: collision detection and ray tracing solve quadratics to find where a moving line meets a circle or a sphere.

  • Optics and communication: parabolic dishes and reflectors focus light and signals, and their cross-section is a quadratic curve.

One formula finds landing points, break-even prices, and collision moments. Anywhere a curve bends back on itself, the Sridharacharya formula is close by.

What Are The Most Common Sridharacharya Formula Mistakes?

These three errors account for most lost marks, and they line up with the calculation mistakes documented in OpenStax's algebra text and in error studies of the quadratic formula.

Getting the sign of bb wrong.

Where it slips in:

When bb is negative, a student substitutes −b-b as if it were still negative, forgetting that the formula already carries a minus sign in front of bb.

Don't do this:

Do not copy bb straight into the −b-b slot without flipping its sign.

The correct way:

Write aa, bb, cc with their signs first, then substitute. For 2x2−x−3=02x^2 - x - 3 = 0, b=−1b = -1, so −b=−(−1)=+1-b = -(-1) = +1, not −1-1.

Dividing only part of the numerator by 2a2a.

Where it slips in:

After reaching x=4±232x = \frac{4 \pm 2\sqrt{3}}{2}, a student divides only the first term and writes 2±232 \pm 2\sqrt{3}, cancelling the 22 with just one piece of the numerator.

Don't do this:

Do not simplify one term and leave the other untouched.

The correct way:

The whole numerator, both −b-b and the root part, is divided by 2a2a. So 4±232=2±3\frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}, dividing every term.

Mishandling a negative discriminant.

Where it slips in:

When b2−4acb^2 - 4ac comes out negative, a student either takes the square root as if the number were positive, or stops and writes "no solution" without naming the complex roots.

Don't do this:

Do not take a real square root of a negative number, and do not read "no real roots" as "no roots at all."

The correct way:

Check the discriminant first. If D<0D < 0, state that there are no real roots and two complex conjugate roots, and only take the square root of a non-negative number.

Practice Problems On The Sridharacharya Formula

Solve each with the formula, checking the discriminant before you substitute. Answers follow each line.

  1. Solve x2−4x−5=0x^2 - 4x - 5 = 0.
    (Answer: D=36D = 36, so x=5x = 5 or x=−1x = -1.)

  2. Solve x2+4x+3=0x^2 + 4x + 3 = 0.
    (Answer: D=4D = 4, so x=−1x = -1 or x=−3x = -3.)

  3. Solve 3x2−5x+2=03x^2 - 5x + 2 = 0.
    (Answer: D=1D = 1, so x=1x = 1 or x=23x = \frac{2}{3}.)

  4. Solve 2x2+5x−3=02x^2 + 5x - 3 = 0.
    (Answer: D=49D = 49, so x=12x = \frac{1}{2} or x=−3x = -3.)

  5. Solve x2−2x−1=0x^2 - 2x - 1 = 0.
    (Answer: D=8D = 8, so x=1+2x = 1 + \sqrt{2} or x=1−2x = 1 - \sqrt{2}.)

  6. Without solving, describe the roots of x2+2x+5=0x^2 + 2x + 5 = 0.
    (Answer: D=4−20=−16<0D = 4 - 20 = -16 < 0, so no real roots; two complex conjugate roots.)

Where Should You Go Next After The Sridharacharya Formula?

The Sridharacharya formula opens onto the wider world of quadratics, and a few natural doors lead out from here.

  1. Quadratic equations. The full family of ax2+bx+c=0ax^2 + bx + c = 0, the different methods for solving them, and when to reach for each.

  2. The discriminant. Go deeper on b2−4acb^2 - 4ac and learn to read the roots of any quadratic before solving it.

  3. Completing the square. The method underneath the formula, useful on its own for vertex form and for graphing parabolas.

If your child is meeting quadratics for the first time, a live Bhanzu trainer builds the formula up from completing the square, so it is understood rather than memorised, in the Bhanzu algebra program.

Was this article helpful?

Your feedback helps us write better content

Frequently Asked Questions

Is the Sridharacharya formula the same as the quadratic formula?
Yes. They are two names for the same result, x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. India often names it after Sridharacharya, who gave the method; most textbooks elsewhere call it the quadratic formula.
Why can't aa be zero in the Sridharacharya formula?
If a=0a = 0, the x2x^2 term disappears and the equation is no longer quadratic; it becomes a linear equation with one solution. The formula also divides by 2a2a, and dividing by zero is undefined.
What does the discriminant tell you?
The discriminant b2−4acb^2 - 4ac reveals the nature of the roots before you solve. Positive gives two real roots, zero gives one repeated root, and negative gives two complex roots.
What happens when the discriminant is negative?
There are no real roots. Instead there are two complex conjugate roots, which involve −1\sqrt{-1}, and the parabola never crosses the x-axis.
Can every quadratic equation be solved with the Sridharacharya formula?
Yes, as long as it is in standard form ax2+bx+c=0ax^2 + bx + c = 0 with a≠0a \neq 0. Factoring works cleanly only for some quadratics, but the Sridharacharya formula solves all of them, including those with irrational or complex roots.
Which class teaches the Sridharacharya formula?
It appears in Class 10 in India (NCERT Chapter 4, Quadratic Equations) and in high-school algebra in the United States (CCSS HSA-REI.B.4). Students usually meet it right after factoring, as the method that never fails.
✍️ Written By
BT
Bhanzu Team
Content Creator and Editor
Bhanzu’s editorial team, known as Team Bhanzu, is made up of experienced educators, curriculum experts, content strategists, and fact-checkers dedicated to making math simple and engaging for learners worldwide. Every article and resource is carefully researched, thoughtfully structured, and rigorously reviewed to ensure accuracy, clarity, and real-world relevance. We understand that building strong math foundations can raise questions for students and parents alike. That’s why Team Bhanzu focuses on delivering practical insights, concept-driven explanations, and trustworthy guidance-empowering learners to develop confidence, speed, and a lifelong love for mathematics.
Related Articles
Book a FREE Demo ClassBook Now →