-22x=2x^2+20

Solution for -22x=2x^2+20 equation:

-22x=2x^2+20

We move all terms to the left:

-22x-(2x^2+20)=0

We get rid of parentheses

-2x^2-22x-20=0

a = -2; b = -22; c = -20;
Δ = b2-4ac
Δ = -222-4·(-2)·(-20)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-18}{2*-2}=\frac{4}{-4}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+18}{2*-2}=\frac{40}{-4}
=-10
$