f^2=60

Solution for f^2=60 equation:

f^2=60

We move all terms to the left:

f^2-(60)=0

a = 1; b = 0; c = -60;
Δ = b2-4ac
Δ = 02-4·1·(-60)
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{15}}{2*1}=\frac{0-4\sqrt{15}}{2}
=-\frac{4\sqrt{15}}{2}
=-2\sqrt{15}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{15}}{2*1}=\frac{0+4\sqrt{15}}{2}
=\frac{4\sqrt{15}}{2}
=2\sqrt{15}
$