r^2+13=26

Solution for r^2+13=26 equation:

r^2+13=26

We move all terms to the left:

r^2+13-(26)=0

We add all the numbers together, and all the variables

r^2-13=0

a = 1; b = 0; c = -13;
Δ = b2-4ac
Δ = 02-4·1·(-13)
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{13}}{2*1}=\frac{0-2\sqrt{13}}{2}
=-\frac{2\sqrt{13}}{2}
=-\sqrt{13}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{13}}{2*1}=\frac{0+2\sqrt{13}}{2}
=\frac{2\sqrt{13}}{2}
=\sqrt{13}
$