k^2+3k-24=4

Solution for k^2+3k-24=4 equation:

k^2+3k-24=4

We move all terms to the left:

k^2+3k-24-(4)=0

We add all the numbers together, and all the variables

k^2+3k-28=0

a = 1; b = 3; c = -28;
Δ = b2-4ac
Δ = 32-4·1·(-28)
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{-(3)-11}{2*1}=\frac{-14}{2}
=-7
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+11}{2*1}=\frac{8}{2}
=4
$