r^2=116

Solution for r^2=116 equation:

r^2=116

We move all terms to the left:

r^2-(116)=0

a = 1; b = 0; c = -116;
Δ = b2-4ac
Δ = 02-4·1·(-116)
Δ = 464
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{464}=\sqrt{16*29}=\sqrt{16}*\sqrt{29}=4\sqrt{29}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{29}}{2*1}=\frac{0-4\sqrt{29}}{2}
=-\frac{4\sqrt{29}}{2}
=-2\sqrt{29}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{29}}{2*1}=\frac{0+4\sqrt{29}}{2}
=\frac{4\sqrt{29}}{2}
=2\sqrt{29}
$