Finding x.

We simplify $\begin{aligned} \sqrt{5-x} & = \; 5 - x^2 \\ 5 - x & = \; (5 - x^2)^2 \\ x^4 - 10x^2 + x + 20 & = \; 0 \\ (x^2 + x - 5)(x^2 - x - 4) & = \; 0 \end{aligned}$ which gives us two solutions $x \; = \; \tfrac12(1 - \sqrt{17}) = -1.56155 \hspace{2cm} x \; = \; \tfrac12(\sqrt{21}-1) = \boxed{1.79129}$ The other two solutions of the quartic satisfy the equation $5 - x^2 = -\sqrt{5-x}$ , and so should be ignored.

Let's have some fun!

$\begin{aligned} \sqrt{5-x} &= 5-x^2\\ 5-x &= 5^2 - 2(5)(x^2) + x^4\\ 0 &= (5)^2 - (2x^2+1)(5) + (x^4 + x)\\ \text{This is a quadratic in 5. Apply the quadratic formula:}\\ 5 &= \frac{2x^2 + 1 \pm \sqrt{(2x^2+1)^2 - 4(x^4+x)}}{2}\\ 5 &= \frac{2x^2 + 1 \pm \sqrt{4x^2 - 4x + 1}}{2}\\ 5 &= \frac{2x^2 + 1 \pm (2x - 1)}{2} \end{aligned}$

Hence, $x^2 + x - 5 = 0$ or $x^2 -x-4 = 0$ . We can then continue with Mark's solution to find the relevant roots.

Disclaimer: This solution was originally mentioned in a BlackPenRedPen video. Check it out! here