117=(2x+14)(x+3)

Solution for 117=(2x+14)(x+3) equation:

117=(2x+14)(x+3)

We move all terms to the left:

117-((2x+14)(x+3))=0

We multiply parentheses ..

-((+2x^2+6x+14x+42))+117=0


We calculate terms in parentheses: -((+2x^2+6x+14x+42)), so:

(+2x^2+6x+14x+42)

We get rid of parentheses

2x^2+6x+14x+42

We add all the numbers together, and all the variables

2x^2+20x+42

Back to the equation:

-(2x^2+20x+42)


We get rid of parentheses

-2x^2-20x-42+117=0

We add all the numbers together, and all the variables

-2x^2-20x+75=0

a = -2; b = -20; c = +75;
Δ = b2-4ac
Δ = -202-4·(-2)·75
Δ = 1000
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:

$\sqrt{\Delta}=\sqrt{1000}=\sqrt{100*10}=\sqrt{100}*\sqrt{10}=10\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-10\sqrt{10}}{2*-2}=\frac{20-10\sqrt{10}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+10\sqrt{10}}{2*-2}=\frac{20+10\sqrt{10}}{-4}
$