MathTools

Quadratic Equation Solver

Solve ax² + bx + c = 0. Roots are exact (fractions when rational, surd form otherwise), and complex roots are reported too.
Equationtwo distinct real roots

1x² + -5x + 6 = 0

Discriminant

1

D = b² − 4ac, two distinct real roots.

Vertex

(5/2, -1/4)

x = −b/2a, the parabola's turning point.

Roots
x = 3≈ 3
x = 2≈ 2
Factored form

(x − 3)(x − 2)

Step-by-step solution

  1. 1

    Identify the coefficients

    a = 1, b = −5, c = 6.

    x2−5x+6=0a=1,b=−5,c=6\begin{aligned}x^{2} - 5x + 6 &= 0 \\ a &= 1,\quad b = -5,\quad c = 6\end{aligned}
  2. 2

    Compute the discriminant

    The discriminant is D = b² − 4ac.

    Δ=b2−4ac=(−5)2−4⋅1⋅6=1\Delta = b^{2} - 4ac = \left(-5\right)^{2} - 4 \cdot 1 \cdot 6 = 1
    • b² = (−5)² = 25
    • 4ac = 4 · 1 · 6 = 24
    • D = 25 − 24 = 1
  3. 3

    Quadratic formula

    x = (−b ± √(b² − 4ac)) / (2a) = (−b ± √D) / (2a).

    x=−b±b2−4ac2a=5±12\begin{aligned}x &= \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \\ &= \frac{5 \pm \sqrt{1}}{2}\end{aligned}
  4. 4

    Interpret the discriminant

    D > 0: two distinct real roots.

  5. 5

    Substitute the values

    x = (−(−5) ± √(1)) / (2).

    x=5±12x = \frac{5 \pm \sqrt{1}}{2}
    • −b = 5
    • 2a = 2
  6. 6

    Simplify the roots

    √D = √(1) = 1 is rational, giving two rational roots.

    x1=52+12=3x2=52−12=2\begin{aligned}x_{1} &= \frac{5}{2} + \frac{1}{2} = 3 \\ x_{2} &= \frac{5}{2} - \frac{1}{2} = 2\end{aligned}
    • x₁ = 5/2 + 1/2 = 3
    • x₂ = 5/2 − 1/2 = 2
  7. 7

    Factored form

    (x − 3)(x − 2)

    (x−3)(x−2)\left(x - 3\right)\left(x - 2\right)
  8. 8

    Vertex of the parabola

    x = −b / (2a) = 5/2, y = c − b² / (4a) = −1/4.

    (−b2a,  c−b24a)=(52,  −14)\left(\frac{-b}{2a},\; c - \frac{b^{2}}{4a}\right) = \left(\frac{5}{2},\; \frac{-1}{4}\right)
  9. 9

    Solution

    Roots: 3, 2.