X(2x+4)+8=20

Solution for X(2x+4)+8=20 equation:

X(2X+4)+8=20

We move all terms to the left:

X(2X+4)+8-(20)=0

We add all the numbers together, and all the variables

X(2X+4)-12=0

We multiply parentheses

2X^2+4X-12=0

a = 2; b = 4; c = -12;
Δ = b2-4ac
Δ = 42-4·2·(-12)
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{7}}{2*2}=\frac{-4-4\sqrt{7}}{4}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{7}}{2*2}=\frac{-4+4\sqrt{7}}{4}
$