2k(k+3)=6k+18

Solution for 2k(k+3)=6k+18 equation:

2k(k+3)=6k+18

We move all terms to the left:

2k(k+3)-(6k+18)=0

We multiply parentheses

2k^2+6k-(6k+18)=0

We get rid of parentheses

2k^2+6k-6k-18=0

We add all the numbers together, and all the variables

2k^2-18=0

a = 2; b = 0; c = -18;
Δ = b2-4ac
Δ = 02-4·2·(-18)
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12}{2*2}=\frac{-12}{4}
=-3
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*2}=\frac{12}{4}
=3
$