68=2x^2+4

Solution for 68=2x^2+4 equation:

68=2x^2+4

We move all terms to the left:

68-(2x^2+4)=0

We get rid of parentheses

-2x^2-4+68=0

We add all the numbers together, and all the variables

-2x^2+64=0

a = -2; b = 0; c = +64;
Δ = b2-4ac
Δ = 02-4·(-2)·64
Δ = 512
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:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{2}}{2*-2}=\frac{0-16\sqrt{2}}{-4}
=-\frac{16\sqrt{2}}{-4}
=-\frac{4\sqrt{2}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{2}}{2*-2}=\frac{0+16\sqrt{2}}{-4}
=\frac{16\sqrt{2}}{-4}
=\frac{4\sqrt{2}}{-1}
$