K2-31-2k=-6-3k^2-2k

Solution for K2-31-2k=-6-3k^2-2k equation:

2-31-2K=-6-3K^2-2K

We move all terms to the left:

2-31-2K-(-6-3K^2-2K)=0

We add all the numbers together, and all the variables

-(-6-3K^2-2K)-2K-29=0

We get rid of parentheses

3K^2+2K-2K+6-29=0

We add all the numbers together, and all the variables

3K^2-23=0

a = 3; b = 0; c = -23;
Δ = b2-4ac
Δ = 02-4·3·(-23)
Δ = 276
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:

$\sqrt{\Delta}=\sqrt{276}=\sqrt{4*69}=\sqrt{4}*\sqrt{69}=2\sqrt{69}$
$K_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{69}}{2*3}=\frac{0-2\sqrt{69}}{6}
=-\frac{2\sqrt{69}}{6}
=-\frac{\sqrt{69}}{3}
$
$K_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{69}}{2*3}=\frac{0+2\sqrt{69}}{6}
=\frac{2\sqrt{69}}{6}
=\frac{\sqrt{69}}{3}
$