2x^2=21^2

Solution for 2x^2=21^2 equation:

2x^2=21^2

We move all terms to the left:

2x^2-(21^2)=0

We add all the numbers together, and all the variables

2x^2-441=0

a = 2; b = 0; c = -441;
Δ = b2-4ac
Δ = 02-4·2·(-441)
Δ = 3528
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:

$\sqrt{\Delta}=\sqrt{3528}=\sqrt{1764*2}=\sqrt{1764}*\sqrt{2}=42\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-42\sqrt{2}}{2*2}=\frac{0-42\sqrt{2}}{4}
=-\frac{42\sqrt{2}}{4}
=-\frac{21\sqrt{2}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+42\sqrt{2}}{2*2}=\frac{0+42\sqrt{2}}{4}
=\frac{42\sqrt{2}}{4}
=\frac{21\sqrt{2}}{2}
$