How do I solve a quadratic for $k$ using the discriminant Picture of the question

Given that the equation $$kx^2-2x+3-2k=0$$ has equal roots, find the possible values of the constant $k$.

So far I have only attempted to factor $-2$ which I'm not sure is even right. I understand I need to use the discriminant later on in the Question. Completely lost, any guidance would be helpful. If you post the solution please show it step by step.
Thanks!
 A: The discriminant of a quadratic $ax^2+bx+c$ is given by $\Delta=b^2-4ac.$ Moreover the quadratic has equal roots when $\Delta=0$.
In this case the quadratic is $kx^2-2x+3-2k$, so $b=-2$, $a=k$ and $c=3-2k$.
Can you end it?
Also by the quadratic equation we have $$x=\frac{2\pm\sqrt{\Delta}}{2k}=\frac{2\pm\sqrt{4-4k(3-2k)}}{2k}$$
$$=\frac{2\pm\sqrt{8k^2-12k+4}}{2k}$$
$$=\frac{2\pm2\sqrt{2k^2-3k+1}}{2k}=\frac{1\pm\sqrt{2k^2-3k+1}}{k}$$
Since you want equal roots (a repeated root), we solve $2k^2-3k+1=0$ for $k$.
A: Once the equation has two equal roots, it means that the discriminant is equal to zero, i.e. $\Delta=0$
For $kx^2-2x+3-2k=0$
$\Delta = b^2 - 4ac \implies \Delta=4-4k(3-2k) \implies \Delta=4-12k+8k^2$
$$0=4-12k+8k^2 \implies 0=4(2k^2-3k+1) \implies 0=4(2k-1)(k-1)$$
$\therefore k=\dfrac{1}{2}; k=1 $
A: It is said equal roots, so $\sqrt{b^2-4ac} = 0$
\begin{align*}
\sqrt{4-4\cdot k\cdot (3-2k)} & = 0\\
2\sqrt{1-3k + 2k^2} &=0 \\
\end{align*}
Now, solve for $2k^2-3k+1 = 0$ which when solved using $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, you will get $k\in\{1,\frac{1}{2}\}$.
A: $$kx^2-2x+3-2k=0$$
Since this equation has equal roots, therefore discriminant is $0$.
$$\Delta=0$$
$$b^2-4ac=0$$
$$2^2-4(3-2k)(k)=0$$
$$1=(3-2k)(k)$$
$$2k^2-3k+1=0$$
$$k=\frac{1}{2},1.$$
