The equation √(3x² + x + 5) = x, where x is real, has no solution. exactly one solution. exactly two solutions. exactly four solutions.

Now, 3x² + x + 5 ≥ 0. This is because b² - 4ac = 1 - 4 × 5 × 3 = -59 (roots are imaginary)
3x² + x + 5 = (x)² = x² + 9
2x² + 7x - 8 = 0
2x² + 8x - x - 8 = 0
2x(x + 4) - (x + 4) = 0
(2x - 1)(x + 4) = 0
x = 1/2 and -4
but x ≥ 3
∴ No solution.