8x+12=2x^2

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

8x+12=2x^2

We move all terms to the left:

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

determiningTheFunctionDomain
-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}
$