20=(x-2)(2x+2)

Solution for 20=(x-2)(2x+2) equation:

20=(x-2)(2x+2)

We move all terms to the left:

20-((x-2)(2x+2))=0

We multiply parentheses ..

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


We calculate terms in parentheses: -((+2x^2+2x-4x-4)), so:

(+2x^2+2x-4x-4)

We get rid of parentheses

2x^2+2x-4x-4

We add all the numbers together, and all the variables

2x^2-2x-4

Back to the equation:

-(2x^2-2x-4)


We get rid of parentheses

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

We add all the numbers together, and all the variables

-2x^2+2x+24=0

a = -2; b = 2; c = +24;
Δ = b2-4ac
Δ = 22-4·(-2)·24
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{-(2)-14}{2*-2}=\frac{-16}{-4}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+14}{2*-2}=\frac{12}{-4}
=-3
$