117=(x+11)(2x+5)

Solution for 117=(x+11)(2x+5) equation:

117=(x+11)(2x+5)

We move all terms to the left:

117-((x+11)(2x+5))=0

We multiply parentheses ..

-((+2x^2+5x+22x+55))+117=0


We calculate terms in parentheses: -((+2x^2+5x+22x+55)), so:

(+2x^2+5x+22x+55)

We get rid of parentheses

2x^2+5x+22x+55

We add all the numbers together, and all the variables

2x^2+27x+55

Back to the equation:

-(2x^2+27x+55)


We get rid of parentheses

-2x^2-27x-55+117=0

We add all the numbers together, and all the variables

-2x^2-27x+62=0

a = -2; b = -27; c = +62;
Δ = b2-4ac
Δ = -272-4·(-2)·62
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1225}=35$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-35}{2*-2}=\frac{-8}{-4}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+35}{2*-2}=\frac{62}{-4}
=-15+1/2
$