Answer
$x_{1}=0$ and $x_{2}=3$
Work Step by Step
Given $\sqrt{4-x} = x-2 \longrightarrow (\sqrt{4-x})^2 = (x-2)^2 \longrightarrow$
$4-x = (x-2)^2 \longrightarrow 4-x = x^2 - 4x + 4 \longrightarrow$
$x^2 - 4x +x + 4 -4 = 0 \longrightarrow x^2 -3x = 0$
$a=1, \ b=-3, \ c=0$
Using the quadratic formula $\dfrac{-b \pm \sqrt{b^2-4ac}}{2a} , $ we have:
$\dfrac{-(-3) \pm \sqrt{(-3)^2-4\times 1\times 0}}{2\times 1} = \dfrac{3 \pm \sqrt{9-0}}{2} = \dfrac{3 \pm \sqrt{9}}{2} = \dfrac{3 \pm 3}{2}$
Therefore we have that $x_{1}= \dfrac{3 + 3}{2}= \dfrac{6}{2} = 3$ and $x_{2}= \dfrac{3 - 3}{2}= \dfrac{0}{2}= 0$
