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

Solution for -8-(k^2+2k-8)=0 equation:

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

We get rid of parentheses

-k^2-2k+8-8=0

We add all the numbers together, and all the variables

-1k^2-2k=0

a = -1; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·(-1)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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