-18x=2x^2+40

Solution for -18x=2x^2+40 equation:

-18x=2x^2+40

We move all terms to the left:

-18x-(2x^2+40)=0

We get rid of parentheses

-2x^2-18x-40=0

a = -2; b = -18; c = -40;
Δ = b2-4ac
Δ = -182-4·(-2)·(-40)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2}{2*-2}=\frac{16}{-4}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2}{2*-2}=\frac{20}{-4}
=-5
$