(x-5)(x-8)=40

Solution for (x-5)(x-8)=40 equation:

(x-5)(x-8)=40

We move all terms to the left:

(x-5)(x-8)-(40)=0

We multiply parentheses ..

(+x^2-8x-5x+40)-40=0

We get rid of parentheses

x^2-8x-5x+40-40=0

We add all the numbers together, and all the variables

x^2-13x=0

a = 1; b = -13; c = 0;
Δ = b2-4ac
Δ = -132-4·1·0
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-13}{2*1}=\frac{0}{2}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+13}{2*1}=\frac{26}{2}
=13
$