Quadratic Equation Solver

Q: What is the formula to solve a quadratic equation?

x = (−b ± √(b² − 4ac)) / 2a, called the quadratic formula or Sridharacharya’s formula. b² − 4ac is the discriminant D.

Q: How do I find the nature of the roots?

Work out D = b² − 4ac. D > 0: two distinct real roots; D = 0: two equal real roots; D 0 and is a perfect square, the roots are rational.

Enter a, b and c of ax² + bx + c = 0 to get the discriminant D = b² − 4ac, the nature of the roots and both roots — as fractions, exact surds like (3 ± √5)/2 or complex numbers like −1 ± 2i, with decimals. When the roots are rational you also get the factorisation by splitting the middle term, exactly as in NCERT Class 10.

Standard form: ax² + bx + c = 0

a (x²)

b (x)

Whole numbers, decimals (0.5) or fractions (1/2). Use 0 for a missing term.

2x² − 7x + 3 = 0

x₁

x₂

1/2

≈ 3 and 0.5

Discriminant D = b² − 4ac

Nature of roots

Two distinct real roots (rational — D is a perfect square)

Factorisation

(x − 3)(2x − 1)

Sum and product of roots

sum 7/2, product 3/2

Vertex (x, y)

(7/4, −25/8)

Parabola

Opens upwards ∪

Step-by-step solution

Compare 2x² − 7x + 3 = 0 with ax² + bx + c = 0: a = 2, b = −7, c = 3
Discriminant D = b² − 4ac = (−7)² − 4 × 2 × 3 = 49 − 24 = 25
D > 0 → Two distinct real roots (rational — D is a perfect square)
x = (−b ± √D) / 2a = (7 ± √25) / 4
x = (7 + 5) / 4 = 3 or x = (7 − 5) / 4 = 1/2
Factorisation by splitting the middle term: two numbers with sum b = −7 and product ac = 6 are −1 and −6
2x² − 7x + 3 = (x − 3)(2x − 1)
Check: sum of roots = −b/a = 7/2, product of roots = c/a = 3/2
Vertex of y = 2x² − 7x + 3: x = −b/2a = 7/4, y = c − b²/4a = −25/8 (the parabola opens upwards — the vertex is the lowest point)

How to use the quadratic equation solver

Write your equation in the standard form ax² + bx + c = 0 (move every term to one side).
Type a, b and c. Use 0 for a missing term, and decimals or fractions if you have them (0.5 or 1/2).
Read the discriminant, the nature of the roots and the two roots, exact and in decimals.
Open the step-by-step solution for the quadratic formula working, the factorisation and the sum and product check.

Formulas

Standard form	`ax² + bx + c = 0, a ≠ 0`
Discriminant	`D = b² − 4ac`
Quadratic formula (Sridharacharya)	`x = (−b ± √D) / 2a`
Nature of roots	`D > 0: two distinct real · D = 0: two equal real · D < 0: no real roots`
Sum and product of roots	`α + β = −b/a, αβ = c/a`
Vertex of y = ax² + bx + c	`x = −b/2a, y = −D/4a`

Solved examples

Example 1: Solve 2x² − 7x + 3 = 0.

a = 2, b = −7, c = 3
D = 49 − 24 = 25 > 0, a perfect square → rational roots
x = (7 ± 5)/4
Split the middle term: −7x = −6x − x → 2x(x − 3) − 1(x − 3) = (x − 3)(2x − 1)

Answer: x = 3 or x = 1/2

Example 2: Find the nature of the roots of x² − 3x + 1 = 0 and solve.

D = 9 − 4 = 5 > 0, not a perfect square
x = (3 ± √5)/2

Answer: Two distinct irrational roots: x ≈ 2.618 or 0.382

Example 3: Solve x² + 2x + 5 = 0.

D = 4 − 20 = −16 < 0 → no real roots
With i = √−1: √−16 = 4i
x = (−2 ± 4i)/2

Answer: x = −1 ± 2i (complex roots, Class 11)

What the discriminant tells you

The quadratic formula has √D in it, so the sign of D decides everything. If D > 0 there are two different real roots; if D is also a perfect square they are rational and the quadratic factorises neatly. If D = 0 the two roots are equal (the parabola just touches the x-axis). If D < 0 there are no real roots — in Class 10 you write “no real roots”, and in Class 11 you find the complex roots p ± qi.

Board questions often ask “find k so that the roots are equal” — set D = 0 and solve for k.

Factorisation, completing the square or the formula?

Splitting the middle term is quickest when D is a perfect square: find two numbers whose sum is b and product is ac. Completing the square and the quadratic formula always work — the formula is just completing the square done once in general. This solver shows the formula and, when possible, the factorisation, so you can check either method.

🎯 Asked in competitive exams

This topic regularly appears in the quantitative aptitude section of:

IBPS PO & Clerk
SBI PO
SSC CGL
JEE Main

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Frequently asked questions

What is the formula to solve a quadratic equation?

x = (−b ± √(b² − 4ac)) / 2a, called the quadratic formula or Sridharacharya’s formula. b² − 4ac is the discriminant D.

How do I find the nature of the roots?

Work out D = b² − 4ac. D > 0: two distinct real roots; D = 0: two equal real roots; D < 0: no real roots. If D > 0 and is a perfect square, the roots are rational.

What if the discriminant is negative?

There are no real roots — that is the Class 10 answer. Using the imaginary unit i = √−1 (Class 11), the roots are complex: x = −b/2a ± (√−D / 2a) i. The solver shows both.

Can I enter decimal or fraction coefficients?

Yes. 0.5x² − 1.5x + 1 = 0 is first multiplied by 2 to give x² − 3x + 2 = 0, which has the same roots, so the exact working stays in whole numbers.

What are the sum and product of the roots?

For ax² + bx + c = 0, the sum of the roots is −b/a and the product is c/a. They are a quick way to check your answer.