117=(3x+1)(2x+1)

Solution for 117=(3x+1)(2x+1) equation:

117=(3x+1)(2x+1)

We move all terms to the left:

117-((3x+1)(2x+1))=0

We multiply parentheses ..

-((+6x^2+3x+2x+1))+117=0


We calculate terms in parentheses: -((+6x^2+3x+2x+1)), so:

(+6x^2+3x+2x+1)

We get rid of parentheses

6x^2+3x+2x+1

We add all the numbers together, and all the variables

6x^2+5x+1

Back to the equation:

-(6x^2+5x+1)


We get rid of parentheses

-6x^2-5x-1+117=0

We add all the numbers together, and all the variables

-6x^2-5x+116=0

a = -6; b = -5; c = +116;
Δ = b2-4ac
Δ = -52-4·(-6)·116
Δ = 2809
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2809}=53$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-53}{2*-6}=\frac{-48}{-12}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+53}{2*-6}=\frac{58}{-12}
=-4+5/6
$