3k^2=-2k+6

Solution for 3k^2=-2k+6 equation:

3k^2=-2k+6

We move all terms to the left:

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

We get rid of parentheses

3k^2+2k-6=0

a = 3; b = 2; c = -6;
Δ = b2-4ac
Δ = 22-4·3·(-6)
Δ = 76
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:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{19}}{2*3}=\frac{-2-2\sqrt{19}}{6}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{19}}{2*3}=\frac{-2+2\sqrt{19}}{6}
$