k+2=k^2-4

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

k+2=k^2-4

We move all terms to the left:

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

We get rid of parentheses

-k^2+k+4+2=0

We add all the numbers together, and all the variables

-1k^2+k+6=0

a = -1; b = 1; c = +6;
Δ = b2-4ac
Δ = 12-4·(-1)·6
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-5}{2*-1}=\frac{-6}{-2}
=+3
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+5}{2*-1}=\frac{4}{-2}
=-2
$