Question

try to answer the question​

Answer 1

B² −4AC=0

Hence

4(2k−3) ²−4(5k−6)(k−2)=0

4k² −12k+9−(5k ² −16k+12)=0

−k ² +4k−3=0

(k−3)(k−1)=0

Hence

k=3 and k=1.