Finding all real solutions of $x-8\sqrt{x}+7=0$

Let $y=\sqrt{x}$, therefore $y^2=x$

$y^2-8y+7=0$ therefore $(y-7)(y-1)=0$ hence $y=1,7$.

Therefore $x=1,49$. These are indeed the only solutions.

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.

If

$x - 8\sqrt x + 7 = 0, \tag 1$

then

$x + 7 = 8\sqrt x; \tag 2$

then

$x^2 + 14 x + 49 = (x + 7)^2 = (8\sqrt x)^2 = 64x; \tag 3$

thus

$x^2 - 50x + 49 = 0, \tag 4$

which factors as

$(x - 1)(x - 49) = x^2 - 50x + 49 = 0; \tag 5$

thus

$x = 1 \; \text{or} \; x = 49; \tag 6$

it is now a simple matter to check that $1$, $49$ obey (1).