r^2=29

Solution for r^2=29 equation:

r^2=29

We move all terms to the left:

r^2-(29)=0

a = 1; b = 0; c = -29;
Δ = b2-4ac
Δ = 02-4·1·(-29)
Δ = 116
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:

$\sqrt{\Delta}=\sqrt{116}=\sqrt{4*29}=\sqrt{4}*\sqrt{29}=2\sqrt{29}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{29}}{2*1}=\frac{0-2\sqrt{29}}{2}
=-\frac{2\sqrt{29}}{2}
=-\sqrt{29}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{29}}{2*1}=\frac{0+2\sqrt{29}}{2}
=\frac{2\sqrt{29}}{2}
=\sqrt{29}
$