Finding all real solutions of $x-8\sqrt{x}+7=0$ Finding all real solutions of $x-8\sqrt{x}+7=0$.
Man, I tried subtituting $x=y^2$ but IDK things got complicated. What is the best way to figure this out? Thanks!

Isolate the square root, then square both sides:
\begin{align*}
     x-8\sqrt x+7&=0\\
     \implies x+7&=8\sqrt x\\
     \implies (x+7)^2 &= 64 x\\
     \implies x^2 -50x + 49&=0\\
     \implies (x-49)(x-1) &=0\\
     \implies x = 49 \text{ or } x&=1,
\end{align*}
which are the two solutions.
