r2-5=3

Solution for r2-5=3 equation:

r2-5=3

We move all terms to the left:

r2-5-(3)=0

We add all the numbers together, and all the variables

r^2-8=0

a = 1; b = 0; c = -8;
Δ = b2-4ac
Δ = 02-4·1·(-8)
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{2}}{2*1}=\frac{0-4\sqrt{2}}{2}
=-\frac{4\sqrt{2}}{2}
=-2\sqrt{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{2}}{2*1}=\frac{0+4\sqrt{2}}{2}
=\frac{4\sqrt{2}}{2}
=2\sqrt{2}
$