Problem Statement
Find all real roots of $\sqrt{5 - x} = 5 - x^2$.
Tutorial 1
Square both sides. After simplification, we have
$$ x^4 - 10x^2 + x + 20 = 0. \tag*{} $$
Of course, you can assume $\gdef\mydef{\gdef}\mydef\s#1{\left(#1\right)}\mydef\m#1{\left[#1\right]}\mydef\b#1{\left\{#1\right\}}\mydef\abs#1{\left\lvert#1\right\rvert}\s{x^2 + ax + b}\s{x^2 + cx + d} = x^4 - 10x^2 + x + 20$. Then, solve for $a$, $b$, $c$ and $d$.
The problem becomes $\s{x^2 - x - 4}\s{x^2 + x - 5} = 0$.
Tutorial 2
Square the both sides.
$$ \begin{aligned} 5 - x &= 5^2 - 5 \cdot 2 x^2 + x^4 \\ 0 &= 5^2 + 5^1 \s{-2x^2 - 1} + 5^0 \s{x^4 + x}. \end{aligned} \tag*{} $$
We can treat it as a quadratic equation in $5$, so apply the quadratic formula:
$$ \begin{aligned} 5 &= \frac{2x^2 + 1 \pm \sqrt{\left(-2x^2 - 1\right)^2 - 4\left(x^4 + x\right)}}2 \\ &= \frac{2x^2 + 1 \pm \sqrt{4x^2 - 4x + 1}}2 \\ &= \frac{2x^2 + 1 \pm \left(2x - 1\right)}2. \end{aligned} \tag*{} $$
Then, after solving the two quadratic equations, you will get the answers.
Tutorial 3
$$ \begin{aligned} \sqrt{5 - x} &= 5 - x^2 \\ x^2 &= 5 - \sqrt{5 - x} \\ x &= \sqrt{5 - \sqrt{5 - x}} \\ x &= \sqrt{5 - \sqrt{5 - \sqrt{5 - \sqrt{5 - \ddots}}}}. \end{aligned} \tag*{} $$
Hence we know that $x = \sqrt{5 - x}$ can give us two solutions. Use long division to find the other quadratic factor and solve it.
Tutorial 4
Note that $\sqrt{5 - x}$ is the inverse function of $5 - x^2$.
Quick check:
$$ \begin{aligned} x &= 5 - y^2 \\ y^2 &= 5 - x \\ y &= \pm\sqrt{5 - x}. \end{aligned} \tag*{} $$
We know that the original function and the inverse function are symmetrical about the line $y = x$.
Hence, their intersections must lie on the line $y = x$, so solving $x = \sqrt{5 - x}$ or $x = 5 - x^2$ gives us two solutions. Use long division to find the other quadratic factor and solve it.