x+x(x+12)+(3x-9)=117

Solution for x+x(x+12)+(3x-9)=117 equation:

x+x(x+12)+(3x-9)=117

We move all terms to the left:

x+x(x+12)+(3x-9)-(117)=0

We multiply parentheses

x^2+x+12x+(3x-9)-117=0

We get rid of parentheses

x^2+x+12x+3x-9-117=0

We add all the numbers together, and all the variables

x^2+16x-126=0

a = 1; b = 16; c = -126;
Δ = b2-4ac
Δ = 162-4·1·(-126)
Δ = 760
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:

$\sqrt{\Delta}=\sqrt{760}=\sqrt{4*190}=\sqrt{4}*\sqrt{190}=2\sqrt{190}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{190}}{2*1}=\frac{-16-2\sqrt{190}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{190}}{2*1}=\frac{-16+2\sqrt{190}}{2}
$