108=(2x+4)(x+5)

Solution for 108=(2x+4)(x+5) equation:

108=(2x+4)(x+5)

We move all terms to the left:

108-((2x+4)(x+5))=0

We multiply parentheses ..

-((+2x^2+10x+4x+20))+108=0


We calculate terms in parentheses: -((+2x^2+10x+4x+20)), so:

(+2x^2+10x+4x+20)

We get rid of parentheses

2x^2+10x+4x+20

We add all the numbers together, and all the variables

2x^2+14x+20

Back to the equation:

-(2x^2+14x+20)


We get rid of parentheses

-2x^2-14x-20+108=0

We add all the numbers together, and all the variables

-2x^2-14x+88=0

a = -2; b = -14; c = +88;
Δ = b2-4ac
Δ = -142-4·(-2)·88
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{-(-14)-30}{2*-2}=\frac{-16}{-4}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+30}{2*-2}=\frac{44}{-4}
=-11
$