(x-5)(x-40)=128

Solution for (x-5)(x-40)=128 equation:

(x-5)(x-40)=128

We move all terms to the left:

(x-5)(x-40)-(128)=0

We multiply parentheses ..

(+x^2-40x-5x+200)-128=0

We get rid of parentheses

x^2-40x-5x+200-128=0

We add all the numbers together, and all the variables

x^2-45x+72=0

a = 1; b = -45; c = +72;
Δ = b2-4ac
Δ = -452-4·1·72
Δ = 1737
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1737}=\sqrt{9*193}=\sqrt{9}*\sqrt{193}=3\sqrt{193}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-45)-3\sqrt{193}}{2*1}=\frac{45-3\sqrt{193}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-45)+3\sqrt{193}}{2*1}=\frac{45+3\sqrt{193}}{2}
$