180=2x^2+19x

Solution for 180=2x^2+19x equation:

180=2x^2+19x

We move all terms to the left:

180-(2x^2+19x)=0

We get rid of parentheses

-2x^2-19x+180=0

a = -2; b = -19; c = +180;
Δ = b2-4ac
Δ = -192-4·(-2)·180
Δ = 1801
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{-(-19)-\sqrt{1801}}{2*-2}=\frac{19-\sqrt{1801}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{1801}}{2*-2}=\frac{19+\sqrt{1801}}{-4}
$