(x-8)(x+10)=40

Solution for (x-8)(x+10)=40 equation:

(x-8)(x+10)=40

We move all terms to the left:

(x-8)(x+10)-(40)=0

We multiply parentheses ..

(+x^2+10x-8x-80)-40=0

We get rid of parentheses

x^2+10x-8x-80-40=0

We add all the numbers together, and all the variables

x^2+2x-120=0

a = 1; b = 2; c = -120;
Δ = b2-4ac
Δ = 22-4·1·(-120)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-22}{2*1}=\frac{-24}{2}
=-12
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+22}{2*1}=\frac{20}{2}
=10
$