3k^2=-64+32k

Solution for 3k^2=-64+32k equation:

3k^2=-64+32k

We move all terms to the left:

3k^2-(-64+32k)=0

We add all the numbers together, and all the variables

3k^2-(32k-64)=0

We get rid of parentheses

3k^2-32k+64=0

a = 3; b = -32; c = +64;
Δ = b2-4ac
Δ = -322-4·3·64
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-16}{2*3}=\frac{16}{6}
=2+2/3
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+16}{2*3}=\frac{48}{6}
=8
$