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

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

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

We move all terms to the left:

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

We get rid of parentheses

k^2+3k^2+2k-3k+6-31=0

We add all the numbers together, and all the variables

4k^2-1k-25=0

a = 4; b = -1; c = -25;
Δ = b2-4ac
Δ = -12-4·4·(-25)
Δ = 401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{401}}{2*4}=\frac{1-\sqrt{401}}{8}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{401}}{2*4}=\frac{1+\sqrt{401}}{8}
$