(x-8)(x+1)=-20

Solution for (x-8)(x+1)=-20 equation:

(x-8)(x+1)=-20

We move all terms to the left:

(x-8)(x+1)-(-20)=0

We add all the numbers together, and all the variables

(x-8)(x+1)+20=0

We multiply parentheses ..

(+x^2+x-8x-8)+20=0

We get rid of parentheses

x^2+x-8x-8+20=0

We add all the numbers together, and all the variables

x^2-7x+12=0

a = 1; b = -7; c = +12;
Δ = b2-4ac
Δ = -72-4·1·12
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-1}{2*1}=\frac{6}{2}
=3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+1}{2*1}=\frac{8}{2}
=4
$