2x^2=22^2

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

2x^2=22^2

We move all terms to the left:

2x^2-(22^2)=0

We add all the numbers together, and all the variables

2x^2-484=0

a = 2; b = 0; c = -484;
Δ = b2-4ac
Δ = 02-4·2·(-484)
Δ = 3872
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:

$\sqrt{\Delta}=\sqrt{3872}=\sqrt{1936*2}=\sqrt{1936}*\sqrt{2}=44\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-44\sqrt{2}}{2*2}=\frac{0-44\sqrt{2}}{4}
=-\frac{44\sqrt{2}}{4}
=-11\sqrt{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+44\sqrt{2}}{2*2}=\frac{0+44\sqrt{2}}{4}
=\frac{44\sqrt{2}}{4}
=11\sqrt{2}
$