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Quadratic Equation Solver

Solve a x²+b x+c=0 with the quadratic formula. The discriminant detects real, double, and complex roots, and the vertex, axis, and parabola graph are shown.

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Input

Enter the coefficients a, b and c of the quadratic equation a x²+b x+c=0. The two roots are found with the quadratic formula.

The equation is a x²+b x+c=0. The coefficient a must not be zero.

Result

Two distinct real roots

Root x₁

2

Root x₂

1

y=a x²+b x+c

Discriminant D

1

Axis of symmetry

x = 1.5

Vertex

( 1.5 , -0.25 )


The roots are x = ( -b ± sqrt(D) ) / 2a with discriminant D = b² - 4ac. When D is negative the roots are complex.

How it works

  • The quadratic formula x = ( -b ± sqrt(D) ) / 2a is used, with discriminant D = b² - 4ac classifying the roots.
  • When D is positive there are two distinct real roots, when D is zero there is a double root, and when D is negative there are conjugate complex roots.
  • The parabola vertex is ( -b / 2a , c - b² / 4a ) and the axis of symmetry is x = -b / 2a.
  • When the coefficient a is zero it is not a quadratic equation, so it cannot be solved.

Frequently asked questions

How is a quadratic equation solved?
With the quadratic formula x = ( -b ± √D ) / 2a, where the discriminant D = b² − 4ac determines the type of roots. If the coefficient a is 0 it is not a quadratic, so it cannot be solved.
How does the discriminant tell the type of roots?
When D = b² − 4ac is positive there are two distinct real roots, when it is 0 there is a repeated (double) root, and when it is negative there are two conjugate complex roots.
Does it also show the vertex and axis of symmetry?
Yes. The parabola's vertex is ( -b / 2a , c - b² / 4a ) and its axis of symmetry is x = -b / 2a, shown together with the graph.

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