2x^2=8x+12=0

Solution for 2x^2=8x+12=0 equation:

2x^2=8x+12=0

We move all terms to the left:

2x^2-(8x+12)=0

We get rid of parentheses

2x^2-8x-12=0

a = 2; b = -8; c = -12;
Δ = b2-4ac
Δ = -82-4·2·(-12)
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{10}}{2*2}=\frac{8-4\sqrt{10}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{10}}{2*2}=\frac{8+4\sqrt{10}}{4}
$