23328=f^2

Solution for 23328=f^2 equation:

23328=f^2

We move all terms to the left:

23328-(f^2)=0

We add all the numbers together, and all the variables

-1f^2+23328=0

a = -1; b = 0; c = +23328;
Δ = b2-4ac
Δ = 02-4·(-1)·23328
Δ = 93312
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{93312}=\sqrt{46656*2}=\sqrt{46656}*\sqrt{2}=216\sqrt{2}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-216\sqrt{2}}{2*-1}=\frac{0-216\sqrt{2}}{-2}
=-\frac{216\sqrt{2}}{-2}
=-\frac{108\sqrt{2}}{-1}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+216\sqrt{2}}{2*-1}=\frac{0+216\sqrt{2}}{-2}
=\frac{216\sqrt{2}}{-2}
=\frac{108\sqrt{2}}{-1}
$