3k*3k+3k=48

Solution for 3k*3k+3k=48 equation:

3k*3k+3k=48

We move all terms to the left:

3k*3k+3k-(48)=0

We add all the numbers together, and all the variables

3k+3k*3k-48=0

Wy multiply elements

9k^2+3k-48=0

a = 9; b = 3; c = -48;
Δ = b2-4ac
Δ = 32-4·9·(-48)
Δ = 1737
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:

$\sqrt{\Delta}=\sqrt{1737}=\sqrt{9*193}=\sqrt{9}*\sqrt{193}=3\sqrt{193}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{193}}{2*9}=\frac{-3-3\sqrt{193}}{18}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{193}}{2*9}=\frac{-3+3\sqrt{193}}{18}
$