(k+1)(k-8)=0

Solution for (k+1)(k-8)=0 equation:

(k+1)(k-8)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

k^2-7k-8=0

a = 1; b = -7; c = -8;
Δ = b2-4ac
Δ = -72-4·1·(-8)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-9}{2*1}=\frac{-2}{2}
=-1
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+9}{2*1}=\frac{16}{2}
=8
$