Question: Solve: $2x +1 = \sqrt{4x²+3x +6}$

Solution:
Given,

$2x +1 = \sqrt{4x²+3x +6}$

[ squaring both sides ]

$or, (2x +1)² = (\sqrt{4x²+3x+6})²$

$or, (2x)² + 2×2x×1 +1² = 4x²+3x+6$

$or, 4x² +4x +1 = 4x² + 3x +6$

$or, 4x²-4x²+4x -3x = 6-1$

$\therefore x = 5$