-2x^2+20=12x

Solution for -2x^2+20=12x equation:

-2x^2+20=12x

We move all terms to the left:

-2x^2+20-(12x)=0

a = -2; b = -12; c = +20;
Δ = b2-4ac
Δ = -122-4·(-2)·20
Δ = 304
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:

$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{19}}{2*-2}=\frac{12-4\sqrt{19}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{19}}{2*-2}=\frac{12+4\sqrt{19}}{-4}
$