Quadratic equations appear from Class 10 onwards and in every engineering entrance exam. Enter the coefficients a, b and c to get the roots, the discriminant, the vertex and a line-by-line solution using the quadratic formula.
Rearrange your equation into the form ax² + bx + c = 0.
Enter a, b and c, including negative signs.
Read the roots and follow the working.
Roots: x = (−b ± √(b² − 4ac)) ÷ 2a
Discriminant: D = b² − 4ac
Vertex: (−b ÷ 2a, c − b² ÷ 4a)
The quadratic equation calculator solves any equation of the form ax² + bx + c = 0. You enter the three coefficients, and it returns the roots, the discriminant D = b² − 4ac and the vertex of the parabola, with every step of the quadratic formula written out. It handles two real roots, one repeated root and complex roots.
Quadratics start in Class 10 and keep appearing in Class 11 complex numbers, JEE, physics and economics. The formula x = (−b ± √D) ÷ 2a is often called Sridharacharya's formula in Indian textbooks, after the mathematician whose method of completing the square is quoted by Bhaskara II. Real problems lead to quadratics whenever something is squared: the height of a thrown ball over time, the area of a plot with a fixed border, or profit when price affects demand.
1. Rearrange the equation into the standard form ax² + bx + c = 0, with everything on one side.
2. Read off a, b and c, keeping their signs.
3. Compute the discriminant D = b² − 4ac.
4. If D > 0, the roots are x = (−b + √D) ÷ 2a and x = (−b − √D) ÷ 2a.
5. If D = 0, there is one repeated root x = −b ÷ 2a. If D < 0, the roots are complex: x = −b ÷ 2a ± (√(−D) ÷ 2|a|) i.
6. Find the vertex at x = −b ÷ 2a and y = c − b² ÷ 4a.
7. Check by substituting each root back into the equation, or by using sum = −b/a and product = c/a.
Divide ax² + bx + c = 0 by a to get x² + (b/a)x = −c/a. Add (b/2a)² to both sides so the left becomes a perfect square: (x + b/2a)² = (b² − 4ac) ÷ 4a². Take square roots: x + b/2a = ±√(b² − 4ac) ÷ 2a. Subtracting b/2a gives x = (−b ± √(b² − 4ac)) ÷ 2a. The expression under the root, the discriminant, is what decides everything about the roots.
The graph of y = ax² + bx + c is a parabola, and its roots are where it crosses the x-axis. If D > 0 it crosses twice, giving two distinct real roots. If D = 0 it just touches the axis at its vertex, giving one repeated root. If D < 0 it stays entirely above or below the axis, and the roots are a pair of complex conjugates p ± qi. When D is a perfect square and a, b, c are integers, the roots are rational and the quadratic factorises neatly.
The parabola is symmetric about the line x = −b/2a, which is also the average of the two roots. Substituting this x gives the vertex height c − b²/4a, the maximum value if a < 0 and the minimum if a > 0. Expanding a(x − α)(x − β) shows that α + β = −b/a and αβ = c/a, relations named after Viète. If a is zero the equation is linear, and the calculator solves bx + c = 0 instead.
Nikhil's Class 10 worksheet asks him to solve 3x² − 5x − 12 = 0 and state the nature of its roots.
Discriminant: D = b² − 4ac D = (-5)² − 4 × 3 × -12 = 169
D > 0: two different real roots: x = (−b ± √D) ÷ 2a x = (5 ± √169) ÷ 6 = (5 ± 13) ÷ 6 x₁ = 3, x₂ = -1.333333
Vertex (turning point): x = −b ÷ 2a = 0.833333, y = c − b² ÷ 4a = -14.083333
Answer: Roots x₁ = 3, x₂ = -1.333333; Discriminant (D) 169; Vertex (0.8333, -14.0833)
Not moving every term to one side first, so c has the wrong sign.
Dropping the minus sign on b. With b = −5, −b is +5.
Squaring a negative b wrongly. (−5)² is 25, not −25.
Dividing only √D by 2a instead of the whole numerator −b ± √D.
Saying there is no solution when D < 0. There are no real roots, but there are two complex ones.
Solving Class 10 and 11 algebra problems and checking factorisation.
Finding when a projectile reaches a given height in physics.
Working out the dimensions of a plot or frame from its area and border.
Finding break-even points when revenue or cost has a squared term.
Locating the maximum or minimum of a quadratic, such as the best price for profit.
What if a is zero?
Then the equation is linear, bx + c = 0, and the calculator solves it as x = −c ÷ b.
What does a negative discriminant mean?
The parabola does not cross the x-axis, so there are no real roots; the roots are complex numbers.