3x^2+20+x^2+12x=180

Solution for 3x^2+20+x^2+12x=180 equation:

3x^2+20+x^2+12x=180

We move all terms to the left:

3x^2+20+x^2+12x-(180)=0

We add all the numbers together, and all the variables

4x^2+12x-160=0

a = 4; b = 12; c = -160;
Δ = b2-4ac
Δ = 122-4·4·(-160)
Δ = 2704
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2704}=52$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-52}{2*4}=\frac{-64}{8}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+52}{2*4}=\frac{40}{8}
=5
$