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

Solution for (x-2)(x+4)=20 equation:

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

We move all terms to the left:

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

We multiply parentheses ..

(+x^2+4x-2x-8)-20=0

We get rid of parentheses

x^2+4x-2x-8-20=0

We add all the numbers together, and all the variables

x^2+2x-28=0

a = 1; b = 2; c = -28;
Δ = b2-4ac
Δ = 22-4·1·(-28)
Δ = 116
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{116}=\sqrt{4*29}=\sqrt{4}*\sqrt{29}=2\sqrt{29}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{29}}{2*1}=\frac{-2-2\sqrt{29}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{29}}{2*1}=\frac{-2+2\sqrt{29}}{2}
$