Solve the following equation by factorization:

$\sqrt{3x^2 + x + 5}$ = (x - 3)

Given,

⇒ $\sqrt{3x^2 + x + 5}$ = (x - 3)

Squaring both sides we get :

⇒ (3x² + x + 5) = (x - 3)²

⇒ 3x² + x + 5 = (x²) + (3²) - 2 × x × 3

⇒ 3x² + x + 5 = x² + 9 - 6x

⇒ 3x² + x + 5 - x² - 9 + 6x = 0

⇒ 2x² + 7x - 4 = 0

⇒ 2x² - x + 8x - 4 = 0

⇒ x(2x - 1) + 4(2x - 1) = 0

⇒ (2x - 1)(x + 4) = 0

⇒ (2x - 1) = 0 or (x + 4) = 0      [Using Zero-product rule]

⇒ 2x = 1 or x = -4

⇒ x = $\dfrac{1}{2}$ or x = -4.

Hence, x = $\Big{-4, \dfrac{1}{2}\Big}$.