k(k+2)=168

Solution for k(k+2)=168 equation:

k(k+2)=168

We move all terms to the left:

k(k+2)-(168)=0

We multiply parentheses

k^2+2k-168=0

a = 1; b = 2; c = -168;
Δ = b2-4ac
Δ = 22-4·1·(-168)
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-26}{2*1}=\frac{-28}{2}
=-14
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+26}{2*1}=\frac{24}{2}
=12
$