2k^2=3k

Solution for 2k^2=3k equation:

2k^2=3k

We move all terms to the left:

2k^2-(3k)=0

a = 2; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·2·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*2}=\frac{0}{4}
=0
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*2}=\frac{6}{4}
=1+1/2
$