x^2+3x+12x-4=180

Solution for x^2+3x+12x-4=180 equation:

x^2+3x+12x-4=180

We move all terms to the left:

x^2+3x+12x-4-(180)=0

We add all the numbers together, and all the variables

x^2+15x-184=0

a = 1; b = 15; c = -184;
Δ = b2-4ac
Δ = 152-4·1·(-184)
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-31}{2*1}=\frac{-46}{2}
=-23
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+31}{2*1}=\frac{16}{2}
=8
$