AlgebraEdge

Algebra Solver

Type the equation or drop in the coefficients — see the discriminant and exact roots, including a + bi form, with steps.

Type the equation (optional)

Any arrangement works: 3x^2 = 19x - 31, x² + 4 = 0, or pasted as 5x 2 + 10 = 13x.

x² − 3x − 4 = 0

x₁ = 4, x₂ = -1

1Write it as ax² + bx + c = 0: x² − 3x − 4 = 0, so a = 1, b = -3, c = -4
2Discriminant Δ = b² − 4ac = (-3)² − 4·1·(-4) = 25
3Apply x = (−b ± √Δ) / (2a) = (3 ± √25) / 2
4Two real roots (or one repeated if Δ = 0): x₁ = 4, x₂ = -1

Roots and discriminants for common textbook quadratics

Every row is the quadratic mode of this solver: type the equation as written or enter a, b and c for ax^2 + bx + c = 0, and it returns the discriminant b^2 - 4ac and the roots in exact form.

Equation	a, b, c to enter	Discriminant	Roots (exact)
x^2 - 3x - 4 = 0	1, -3, -4	25	x = 4 or x = -1
x^2 - 5x + 6 = 0	1, -5, 6	1	x = 3 or x = 2
x^2 - 4 = 0	1, 0, -4	16	x = 2 or x = -2
x^2 - 2x - 8 = 0	1, -2, -8	36	x = 4 or x = -2
2x^2 - 4x - 6 = 0	2, -4, -6	64	x = 3 or x = -1
3x^2 + 5x - 2 = 0	3, 5, -2	49	x = 1/3 or x = -2
6x^2 + 11x - 35 = 0	6, 11, -35	961	x = 5/3 or x = -7/2
x^2 - 2x - 1 = 0	1, -2, -1	8	x = 1 +/- sqrt(2)
x^2 + 2x + 1 = 0	1, 2, 1	0	x = -1 (repeated)
x^2 + x + 1 = 0	1, 1, 1	-3	x = -1/2 +/- (sqrt(3)/2)i
x^2 + 4x + 5 = 0	1, 4, 5	-4	x = -2 +/- i
3x^2 = 19x - 31	3, -19, 31	-11	x = 19/6 +/- (sqrt(11)/6)i

A positive discriminant gives two distinct real roots, zero gives one repeated root, and a negative value gives a conjugate pair of complex roots in a + bi form, as in the last three rows. Roots are exact: reduced fractions, simplified square roots, and complex roots split into a real part -b/2a and an imaginary part (written +/- and sqrt here; the solver shows the symbols and the decimals). Switch to linear mode for ax + b = 0, where the single root is always x = -b/a.

Reading the discriminant

For a quadratic, the discriminant b²−4ac tells you what kind of roots to expect. Positive: two real roots. Zero: one repeated real root. Negative: two complex conjugate roots.

The quadratic formula

x = (−b ± √(b²−4ac)) / 2a. This works for any quadratic equation ax² + bx + c = 0, regardless of whether it factors neatly. The ± means there are usually two solutions — plug in + for one root and − for the other.

Frequently asked questions

How do I solve a quadratic equation?

Three methods: (1) Factoring — if ax² + bx + c factors into (px + q)(rx + s) = 0, set each factor to zero. (2) Quadratic formula — x = (−b ± √(b²−4ac)) / 2a, always works. (3) Completing the square — rewrite as (x + h)² = k, then x = −h ± √k. This calculator uses the quadratic formula and shows every step.

How do I write complex roots in simplest a + bi form?

Move every term to one side, find the discriminant, and split the quadratic formula into a real part and an imaginary part. Example: 3x² = 19x − 31 becomes 3x² − 19x + 31 = 0, so a = 3, b = −19, c = 31 and b² − 4ac = 361 − 372 = −11. The roots are (19 ± √−11)/6 = (19 ± i√11)/6, which in a + bi form is x = 19/6 ± (√11/6)i: reduce the real fraction −b/2a and write the imaginary part as a simplified radical over 2a. Type your equation into the box above and the solver does each of these steps.

What does 'no real solution' mean?

When the discriminant (b²−4ac) is negative, the square root of a negative number doesn't exist in real numbers. The equation has two complex conjugate solutions: x = (−b ± i√|b²−4ac|) / 2a. Graphically, the parabola doesn't cross the x-axis.

How do I solve a linear equation?

For ax + b = 0: subtract b from both sides, then divide by a. x = −b/a. Example: 3x + 12 = 0 → 3x = −12 → x = −4. Linear equations always have exactly one solution (unless a = 0, in which case it's either no solution or infinitely many).