(x^2+112)+(16x+131)=180

Solution for (x^2+112)+(16x+131)=180 equation:

(x^2+112)+(16x+131)=180

We move all terms to the left:

(x^2+112)+(16x+131)-(180)=0

We get rid of parentheses

x^2+16x+112+131-180=0

We add all the numbers together, and all the variables

x^2+16x+63=0

a = 1; b = 16; c = +63;
Δ = b2-4ac
Δ = 162-4·1·63
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2}{2*1}=\frac{-18}{2}
=-9
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2}{2*1}=\frac{-14}{2}
=-7
$