(x^2+80)+(3x+12)=180

Solution for (x^2+80)+(3x+12)=180 equation:

(x^2+80)+(3x+12)=180

We move all terms to the left:

(x^2+80)+(3x+12)-(180)=0

We get rid of parentheses

x^2+3x+80+12-180=0

We add all the numbers together, and all the variables

x^2+3x-88=0

a = 1; b = 3; c = -88;
Δ = b2-4ac
Δ = 32-4·1·(-88)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-19}{2*1}=\frac{-22}{2}
=-11
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+19}{2*1}=\frac{16}{2}
=8
$