We have,
\(\sqrt{6x + 7} - (2x - 7) = 0\)
\(\sqrt{6x + 7} = (2x - 7) \)
On squaring both sides
\((\sqrt{6x + 7} )^2= (2x - 7) ^2\)
⇒ 6x + 7 = 4x2 + 49 − 28x
⇒ 4x2 + 42 − 34x = 0
⇒ 2x2 − 17x + 21 = 0
⇒ 2x2 − 14x − 3x + 21 = 0
⇒ 2x (x−7) − 3 (x − 7) = 0
⇒ (2x − 3) (x − 7) = 0
⇒ x = \(\frac 32\) or 7
∴ x = 7 (as x = 3/2 doesn't satisfy the given equation)
