22=2x^2=+4

Solution for 22=2x^2=+4 equation:

22=2x^2=+4

We move all terms to the left:

22-(2x^2)=0

a = -2; b = 0; c = +22;
Δ = b2-4ac
Δ = 02-4·(-2)·22
Δ = 176
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:

$\sqrt{\Delta}=\sqrt{176}=\sqrt{16*11}=\sqrt{16}*\sqrt{11}=4\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{11}}{2*-2}=\frac{0-4\sqrt{11}}{-4}
=-\frac{4\sqrt{11}}{-4}
=-\frac{\sqrt{11}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{11}}{2*-2}=\frac{0+4\sqrt{11}}{-4}
=\frac{4\sqrt{11}}{-4}
=\frac{\sqrt{11}}{-1}
$