2x^2=16x+12

Solution for 2x^2=16x+12 equation:

2x^2=16x+12

We move all terms to the left:

2x^2-(16x+12)=0

We get rid of parentheses

2x^2-16x-12=0

a = 2; b = -16; c = -12;
Δ = b2-4ac
Δ = -162-4·2·(-12)
Δ = 352
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:

$\sqrt{\Delta}=\sqrt{352}=\sqrt{16*22}=\sqrt{16}*\sqrt{22}=4\sqrt{22}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{22}}{2*2}=\frac{16-4\sqrt{22}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{22}}{2*2}=\frac{16+4\sqrt{22}}{4}
$