-8r+r^2=16

Solution for -8r+r^2=16 equation:

-8r+r^2=16

We move all terms to the left:

-8r+r^2-(16)=0

a = 1; b = -8; c = -16;
Δ = b2-4ac
Δ = -82-4·1·(-16)
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8\sqrt{2}}{2*1}=\frac{8-8\sqrt{2}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8\sqrt{2}}{2*1}=\frac{8+8\sqrt{2}}{2}
$