180=2x^2+3x

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

180=2x^2+3x

We move all terms to the left:

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

We get rid of parentheses

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

a = -2; b = -3; c = +180;
Δ = b2-4ac
Δ = -32-4·(-2)·180
Δ = 1449
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1449}=\sqrt{9*161}=\sqrt{9}*\sqrt{161}=3\sqrt{161}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3\sqrt{161}}{2*-2}=\frac{3-3\sqrt{161}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3\sqrt{161}}{2*-2}=\frac{3+3\sqrt{161}}{-4}
$