(x-5)(2x-8)=0

Solution for (x-5)(2x-8)=0 equation:

(x-5)(2x-8)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2-18x+40=0

a = 2; b = -18; c = +40;
Δ = b2-4ac
Δ = -182-4·2·40
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2}{2*2}=\frac{16}{4}
=4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2}{2*2}=\frac{20}{4}
=5
$