k(k+2)=17

Solution for k(k+2)=17 equation:

k(k+2)=17

We move all terms to the left:

k(k+2)-(17)=0

We multiply parentheses

k^2+2k-17=0

a = 1; b = 2; c = -17;
Δ = b2-4ac
Δ = 22-4·1·(-17)
Δ = 72
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-6\sqrt{2}}{2*1}=\frac{-2-6\sqrt{2}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6\sqrt{2}}{2*1}=\frac{-2+6\sqrt{2}}{2}
$