3x^2+33x=180

Solution for 3x^2+33x=180 equation:

3x^2+33x=180

We move all terms to the left:

3x^2+33x-(180)=0

a = 3; b = 33; c = -180;
Δ = b2-4ac
Δ = 332-4·3·(-180)
Δ = 3249
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3249}=57$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-57}{2*3}=\frac{-90}{6}
=-15
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+57}{2*3}=\frac{24}{6}
=4
$