-r^2+8=0

Solution for -r^2+8=0 equation:

-r^2+8=0

We add all the numbers together, and all the variables

-1r^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}
=-\frac{2\sqrt{2}}{-1}
$
$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}
=\frac{2\sqrt{2}}{-1}
$