3k(k+1)=18

Solution for 3k(k+1)=18 equation:

3k(k+1)=18

We move all terms to the left:

3k(k+1)-(18)=0

We multiply parentheses

3k^2+3k-18=0

a = 3; b = 3; c = -18;
Δ = b2-4ac
Δ = 32-4·3·(-18)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-15}{2*3}=\frac{-18}{6}
=-3
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+15}{2*3}=\frac{12}{6}
=2
$