x+(x+10)(x+5)=108

Solution for x+(x+10)(x+5)=108 equation:

x+(x+10)(x+5)=108

We move all terms to the left:

x+(x+10)(x+5)-(108)=0

We multiply parentheses ..

(+x^2+5x+10x+50)+x-108=0

We get rid of parentheses

x^2+5x+10x+x+50-108=0

We add all the numbers together, and all the variables

x^2+16x-58=0

a = 1; b = 16; c = -58;
Δ = b2-4ac
Δ = 162-4·1·(-58)
Δ = 488
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:

$\sqrt{\Delta}=\sqrt{488}=\sqrt{4*122}=\sqrt{4}*\sqrt{122}=2\sqrt{122}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{122}}{2*1}=\frac{-16-2\sqrt{122}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{122}}{2*1}=\frac{-16+2\sqrt{122}}{2}
$