Answer
$7$.
Work Step by Step
From the question we have
$\Rightarrow \sqrt{x-3}=x-5$
Square both sides.
$\Rightarrow (\sqrt{x-3})^2=(x-5)^2$
Use the special formula $(A-B)^2=A^2-2AB+B^2$
We have $A=x$ and $B=5$
$\Rightarrow x-3=(x)^2-2(x)( 5)+(5)^2$
Simplify.
$\Rightarrow x-3=x^2-10x+25$
Add $-x+3$ to both sides.
$\Rightarrow x-3-x+3=x^2-10x+25-x+3$
Add like terms.
$\Rightarrow 0=x^2-11x+28$
Rewrite the middle term $-11x$ as $-7x-4x$.
$\Rightarrow 0=x^2-7x-4x+28$
Group the terms.
$\Rightarrow 0=(x^2-7x)+(-4x+28)$
Factor out each group.
$\Rightarrow 0=x(x-7)-4(x-7)$
Factor out $(x-7)$.
$\Rightarrow 0=(x-7)(x-4)$
Set each factor equal to zero.
$\Rightarrow x-7=0$ or $x-4=0$
Isolate $x$.
$\Rightarrow x=7$ or $x=4$
Check $x=7$.
$\Rightarrow \sqrt{(7)-3}=7-5$
$\Rightarrow \sqrt{7-3}=2$
$\Rightarrow \sqrt{4}=2$
$\Rightarrow 2=2$ True.
Check $x=4$.
$\Rightarrow \sqrt{(4)-3}=4-5$
$\Rightarrow \sqrt{4-3}=-1$
$\Rightarrow \sqrt{1}=-1$
$\Rightarrow 1=-1$ false.
The solution is $7$.
