6.k^2+4k=96

Solution for 6.k^2+4k=96 equation:

6.k^2+4k=96

We move all terms to the left:

6.k^2+4k-(96)=0

a = 6.; b = 4; c = -96;
Δ = b2-4ac
Δ = 42-4·6.·(-96)
Δ = 2320
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:

$\sqrt{\Delta}=\sqrt{2320}=\sqrt{16*145}=\sqrt{16}*\sqrt{145}=4\sqrt{145}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{145}}{2*6.}=\frac{-4-4\sqrt{145}}{12}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{145}}{2*6.}=\frac{-4+4\sqrt{145}}{12}
$