(x-5)*(2x+4)=0

Solution for (x-5)*(2x+4)=0 equation:

(x-5)(2x+4)=0

We multiply parentheses ..

(+2x^2+4x-10x-20)=0

We get rid of parentheses

2x^2+4x-10x-20=0

We add all the numbers together, and all the variables

2x^2-6x-20=0

a = 2; b = -6; c = -20;
Δ = b2-4ac
Δ = -62-4·2·(-20)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-14}{2*2}=\frac{-8}{4}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+14}{2*2}=\frac{20}{4}
=5
$