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

Solution for (2x+10)(2x-8)=0 equation:

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

We multiply parentheses ..

(+4x^2-16x+20x-80)=0

We get rid of parentheses

4x^2-16x+20x-80=0

We add all the numbers together, and all the variables

4x^2+4x-80=0

a = 4; b = 4; c = -80;
Δ = b2-4ac
Δ = 42-4·4·(-80)
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-36}{2*4}=\frac{-40}{8}
=-5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+36}{2*4}=\frac{32}{8}
=4
$