4+r^2=125

Solution for 4+r^2=125 equation:

4+r^2=125

We move all terms to the left:

4+r^2-(125)=0

We add all the numbers together, and all the variables

r^2-121=0

a = 1; b = 0; c = -121;
Δ = b2-4ac
Δ = 02-4·1·(-121)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-22}{2*1}=\frac{-22}{2}
=-11
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+22}{2*1}=\frac{22}{2}
=11
$