Answer
$x=4$
Work Step by Step
$\sqrt{13-x}=x-1$
$(\sqrt{13-x})^2=(x-1)^2$
$13-x=x^2-2x+1$
$13-x-13+x=x^2-2x+1-13+x$
$0=x^2-x-12$
$0=(x-4)(x+3)$
$0=x-4$
$x=4$
$0=x+3$
$x=-3$
$\sqrt{13-x}=x-1$
$\sqrt{13-4}=4-1$
$\sqrt{9}=3$
$3=3$ (true)
$\sqrt{13--3}=-3-1$
$\sqrt {16}=-4$
$4\ne-4$ (false)
