Enter coefficients a, b, c for ax² + bx + c = 0 and get real or complex roots instantly.
Solving 2x² + 5x − 3 = 0.
Two real roots: x = 0.5 and x = −3.
The sign of b² − 4ac answers the question people usually actually have. Positive means the parabola crosses the horizontal axis twice, zero means it touches it at a single point, and negative means it never reaches it and the solutions are complex. If you only need to know whether real solutions exist — whether a projectile reaches a height, whether two paths intersect — the discriminant is the whole answer and you can stop there.
Two sign traps account for most errors. First, c carries its own sign into the formula, so with c negative the −4ac term becomes an addition and the discriminant grows. Second, the −b in the numerator flips the sign of b, so a positive b produces roots built from a negative number. Substituting your answers back into the original equation takes ten seconds and catches both.
With a = 0 there is no squared term and the equation is linear, not quadratic — bx + c = 0 solves directly as x = −c ÷ b. The formula itself would divide by 2a, which is zero, so the calculator refuses the input rather than returning an infinity.
A negative discriminant does not mean the problem is broken. The two roots are a conjugate pair, sharing the same real part −b ÷ 2a and differing only in the sign of the imaginary part, and the calculator displays both. In a physical problem that usually translates to no real crossing point — the parabola stays entirely above or entirely below the axis — which is often exactly the information you were after.
x = (−b ± √(b² − 4ac)) ÷ 2a, applied to any equation written in the form ax² + bx + c = 0. The plus and minus give the two roots, and the expression under the square root is the discriminant that decides whether those roots are real.
How many real solutions there are. Positive gives two distinct real roots, zero gives one repeated root where the curve touches the axis, and negative gives two complex conjugate roots and no real crossing at all.
Because the equation is then linear rather than quadratic, and the formula divides by 2a. Solve bx + c = 0 directly instead: x = −c ÷ b.
Substitute each root back into the original equation and confirm it evaluates to zero. As a second check, the two roots should sum to −b ÷ a and multiply to c ÷ a — for 2x² + 5x − 3 that is 0.5 + (−3) = −2.5 and 0.5 × (−3) = −1.5, which match.
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