k(k-2)=15

Solution for k(k-2)=15 equation:

k(k-2)=15

We move all terms to the left:

k(k-2)-(15)=0

We multiply parentheses

k^2-2k-15=0

a = 1; b = -2; c = -15;
Δ = b2-4ac
Δ = -22-4·1·(-15)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-8}{2*1}=\frac{-6}{2}
=-3
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+8}{2*1}=\frac{10}{2}
=5
$