3k^2+2k=96

Solution for 3k^2+2k=96 equation:

3k^2+2k=96

We move all terms to the left:

3k^2+2k-(96)=0

a = 3; b = 2; c = -96;
Δ = b2-4ac
Δ = 22-4·3·(-96)
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1156}=34$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-34}{2*3}=\frac{-36}{6}
=-6
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+34}{2*3}=\frac{32}{6}
=5+1/3
$