2(x-2)(x+1)=20

Solution for 2(x-2)(x+1)=20 equation:

2(x-2)(x+1)=20

We move all terms to the left:

2(x-2)(x+1)-(20)=0

We multiply parentheses ..

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

We multiply parentheses

2x^2+2x-4x-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{-12}{4}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+14}{2*2}=\frac{16}{4}
=4
$