How do you solve the quadratic $(x-5)(x+8)=-20$ using any method?

Question

3285 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-16T16:38:02

$x = 1/2(-3+-sqrt(89))$

$-20 = (x-5)(x+8)=x^2+3x-40$

Add $20$ to both ends to get:

$0 = x^2+3x-20$

Multiply by $4$ and complete the square:

$0 = 4x^2+12x-80$

$=(2x)^2+2(3)(2x)+3^2-89$

$=(2x+3)^2-(sqrt(89))^2$

$=(2x+3-sqrt(89))(2x+3+sqrt(89))$

Hence $x = 1/2(-3+-sqrt(89))$

How do you solve the quadratic (x-5)(x+8)=-20(x-5)(x+8)=-20 using any method?