20x=2x^2+12

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

20x=2x^2+12

We move all terms to the left:

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

We get rid of parentheses

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

a = -2; b = 20; c = -12;
Δ = b2-4ac
Δ = 202-4·(-2)·(-12)
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{-(20)-4\sqrt{19}}{2*-2}=\frac{-20-4\sqrt{19}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{19}}{2*-2}=\frac{-20+4\sqrt{19}}{-4}
$