Description for Correct answer:
We have

\( \Large \sqrt{3x^{2}-7x-30}+\sqrt{2x^{2}-7x-5}=x+5 \)

or \( \Large \sqrt{3x^{2}-7x-30}= \left(x+5\right)-\sqrt{2x^{2}-7x-5} \)

on squaring both sides, we get

\( \Large 3x^{2}-7x-30=x^{2}+25+10x+ \left(2x^{2}-7x-5\right) \)

\( \Large -2 \left(x+5\right)\sqrt{2x^{2}-7x-5} => \sqrt{2x^{2}-7x-5}=5 \)

Again squaring

=> \( \Large 2x^{2}-7x-30=0 => x=6 \)