(9x+4)+(x^2+11)+(2x^2+2x-3)=180

Solution for (9x+4)+(x^2+11)+(2x^2+2x-3)=180 equation:

(9x+4)+(x^2+11)+(2x^2+2x-3)=180

We move all terms to the left:

(9x+4)+(x^2+11)+(2x^2+2x-3)-(180)=0

We get rid of parentheses

x^2+2x^2+9x+2x+4+11-3-180=0

We add all the numbers together, and all the variables

3x^2+11x-168=0

a = 3; b = 11; c = -168;
Δ = b2-4ac
Δ = 112-4·3·(-168)
Δ = 2137
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-\sqrt{2137}}{2*3}=\frac{-11-\sqrt{2137}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{2137}}{2*3}=\frac{-11+\sqrt{2137}}{6}
$