2x^2-18x=-18

Solution for 2x^2-18x=-18 equation:

2x^2-18x=-18

We move all terms to the left:

2x^2-18x-(-18)=0

We add all the numbers together, and all the variables

2x^2-18x+18=0

a = 2; b = -18; c = +18;
Δ = b2-4ac
Δ = -182-4·2·18
Δ = 180
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:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6\sqrt{5}}{2*2}=\frac{18-6\sqrt{5}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6\sqrt{5}}{2*2}=\frac{18+6\sqrt{5}}{4}
$