x = 4 is the solution
given [tex]\sqrt{x+12}[/tex] = x ( square both sides )
x + 12 = x² ( rearrange into standard form )
x² - x - 12 = 0 ← in standard form
(x - 4 )(x + 3 ) = 0 ← in factored form
equate each factor to zero and solve for x
x - 4 = 0 ⇒ x = 4
x + 3 = 0 ⇒ x = - 3
substitute these values into the original equation and if both sides are equal then they are the solutions
x = 4 : [tex]\sqrt{4+12}[/tex] = [tex]\sqrt{16}[/tex] = 4 = x ← correct
x = - 3 : [tex]\sqrt{-3+12}[/tex] = [tex]\sqrt{9}[/tex] = 3 ≠ - 3 ← false
the solution is x = 4
