Value of Polinomials (2)

$\begin{aligned} x^2-16\sqrt{x}&=12 \hspace{4mm}\small\color{#3D99F6}\text{(Given)}\\ \text{Adding } 4x+4& \text{ to both sides,}\\ x^2+4x+4&=4x+16\sqrt{x}+16\\ \implies (x+2)^2 &=(2\sqrt{x}+4)^2\\ \implies(x+2)^2-(2\sqrt{x}+4)^2&=0\\ (x+2+2\sqrt{x}+4)(x+2-2\sqrt{x}-4)&=0\\ \implies x&=2\sqrt{x}+2 \text{ or } x=-2\sqrt{x}-6\\\\ \text{However,}\\ \text{Setting } x&=-2\sqrt{x}-6 \\ \implies (\sqrt{x}-1)^2&=-5\\ \text{The above equation has} &\text{ no real solution.}\\\\ \implies (x-2\sqrt{x})&=\color{#EC7300}\boxed{\color{#333333}2}\end{aligned}$

Let $y=\sqrt{x}$ . Then we have $\begin{aligned} y^4 - 16 y &= 12 \\ y^4 - 16y -12 &= 0 \\ (y^2 - 2y - 2)(y^2 + 2y + 6) &= 0 && \color{#3D99F6}\text{Only }y^2-2y-2\text{ has real roots} \\ y^2 - 2y - 2 &= 0 \\ y^2 - 2y &= 2 \end{aligned}$ We could solve for $y$ , but note that $x-2\sqrt{x} = y^2 - 2y = \boxed{2}$