k^2-2k=-3k^2+17

Solution for k^2-2k=-3k^2+17 equation:

k^2-2k=-3k^2+17

We move all terms to the left:

k^2-2k-(-3k^2+17)=0

We get rid of parentheses

k^2+3k^2-2k-17=0

We add all the numbers together, and all the variables

4k^2-2k-17=0

a = 4; b = -2; c = -17;
Δ = b2-4ac
Δ = -22-4·4·(-17)
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{-(-2)-2\sqrt{69}}{2*4}=\frac{2-2\sqrt{69}}{8}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{69}}{2*4}=\frac{2+2\sqrt{69}}{8}
$