(c^2)+8c=8

Solution for (c^2)+8c=8 equation:

(c^2)+8c=8

We move all terms to the left:

(c^2)+8c-(8)=0

a = 1; b = 8; c = -8;
Δ = b2-4ac
Δ = 82-4·1·(-8)
Δ = 96
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{6}}{2*1}=\frac{-8-4\sqrt{6}}{2}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{6}}{2*1}=\frac{-8+4\sqrt{6}}{2}
$