How to Solve a Quadratic Equation

One equation, x2 + 5x - 14 = 0, solved three ways, so the working can be copied for whichever method the question demands.

Start with factorisation. Look for two numbers that multiply to -14 and add to 5. The factor pairs of 14 are 1 x 14 and 2 x 7, and only 7 and 2 differ by 5; because c is negative the signs differ, giving (x + 7)(x - 2) = 0. Setting each bracket to zero gives x = -7 and x = 2.

Check both roots in the original: 49 - 35 - 14 = 0 and 4 + 10 - 14 = 0. Both hold, so the factorisation is correct and the solution is complete.

If no pair appears, use the formula x = (-b +/- sqrt(b2 - 4ac)) / 2a with a = 1, b = 5, c = -14: x = (-5 +/- sqrt(25 + 56)) / 2 = (-5 +/- 9) / 2, giving x = 2 and x = -7. The discriminant being 81, a perfect square, is exactly why factorisation worked.

Completing the square reaches the same pair: x2 + 5x = 14 becomes (x + 5/2)2 = 14 + 25/4 = 81/4, so x + 5/2 = +/- 9/2 and x = 2 or x = -7. For an equation that does not factorise, such as x2 + 6x = 5, the same routine gives (x + 3)2 = 14 and x = -3 +/- sqrt(14), about 0.74 and -6.74.

Choose by the discriminant: if b2 - 4ac is a perfect square, factorise; if not, the formula or completing the square still gets there. The discriminant page shows the three cases in full, and the simultaneous page applies the same one-line-per-operation layout to pairs of equations.

• a = 1, b = 5, c = -14 for the worked equation
• factor pairs of 14: 1 x 14 and 2 x 7
• roots: x = -7 and x = 2
• check lines: 49 - 35 - 14 = 0 and 4 + 10 - 14 = 0

Further reading

• What the Discriminant Tells You
• Solving Simultaneous Equations by Elimination