(k+8)(k-3)=0

Solution for (k+8)(k-3)=0 equation:

(k+8)(k-3)=0

We multiply parentheses ..

(+k^2-3k+8k-24)=0

We get rid of parentheses

k^2-3k+8k-24=0

We add all the numbers together, and all the variables

k^2+5k-24=0

a = 1; b = 5; c = -24;
Δ = b2-4ac
Δ = 52-4·1·(-24)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-11}{2*1}=\frac{-16}{2}
=-8
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+11}{2*1}=\frac{6}{2}
=3
$