36=2x^2+3x

Solution for 36=2x^2+3x equation:

36=2x^2+3x

We move all terms to the left:

36-(2x^2+3x)=0

We get rid of parentheses

-2x^2-3x+36=0

a = -2; b = -3; c = +36;
Δ = b2-4ac
Δ = -32-4·(-2)·36
Δ = 297
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:

$\sqrt{\Delta}=\sqrt{297}=\sqrt{9*33}=\sqrt{9}*\sqrt{33}=3\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3\sqrt{33}}{2*-2}=\frac{3-3\sqrt{33}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3\sqrt{33}}{2*-2}=\frac{3+3\sqrt{33}}{-4}
$