Find the value of $k$ for which the quadratic equation $\displaystyle \left ( k-2 \right )x^{2}+2\left ( 2k-3 \right )x+5k-6=0$ has equal roots

Since the above equation has equal roots

Hence

$B^{2}-4AC=0$

Hence

$4(2k-3)^{2}-4(5k-6)(k-2)=0$

$4k^{2}-12k+9-(5k^{2}-16k+12)=0$

$-k^{2}+4k-3=0$

Or ;

$k^{2}-4k+3=0$

$(k-3)(k-1)=0$

Hence

$k=3$ and $k=1$.