(x+1)(2x-4)=108

Solution for (x+1)(2x-4)=108 equation:

(x+1)(2x-4)=108

We move all terms to the left:

(x+1)(2x-4)-(108)=0

We multiply parentheses ..

(+2x^2-4x+2x-4)-108=0

We get rid of parentheses

2x^2-4x+2x-4-108=0

We add all the numbers together, and all the variables

2x^2-2x-112=0

a = 2; b = -2; c = -112;
Δ = b2-4ac
Δ = -22-4·2·(-112)
Δ = 900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{900}=30$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-30}{2*2}=\frac{-28}{4}
=-7
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+30}{2*2}=\frac{32}{4}
=8
$