3a^2=11=21+a^2

Solution for 3a^2=11=21+a^2 equation:

3a^2=11=21+a^2

We move all terms to the left:

3a^2-(11)=0

a = 3; b = 0; c = -11;
Δ = b2-4ac
Δ = 02-4·3·(-11)
Δ = 132
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{33}}{2*3}=\frac{0-2\sqrt{33}}{6}
=-\frac{2\sqrt{33}}{6}
=-\frac{\sqrt{33}}{3}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{33}}{2*3}=\frac{0+2\sqrt{33}}{6}
=\frac{2\sqrt{33}}{6}
=\frac{\sqrt{33}}{3}
$