k2-31-2k=-6-3k2-2k

Solution for k2-31-2k=-6-3k2-2k equation:

k2-31-2k=-6-3k^2-2k

We move all terms to the left:

k2-31-2k-(-6-3k^2-2k)=0

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

4k^2-25=0

a = 4; b = 0; c = -25;
Δ = b2-4ac
Δ = 02-4·4·(-25)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20}{2*4}=\frac{-20}{8}
=-2+1/2
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20}{2*4}=\frac{20}{8}
=2+1/2
$