k^2+k=6

Solution for k^2+k=6 equation:

k^2+k=6

We move all terms to the left:

k^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
$