0=-2x^2+36x

Solution for 0=-2x^2+36x equation:

0=-2x^2+36x

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

2x^2-36x=0

a = 2; b = -36; c = 0;
Δ = b2-4ac
Δ = -362-4·2·0
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-36}{2*2}=\frac{0}{4}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+36}{2*2}=\frac{72}{4}
=18
$