Answer
$x=2$
Work Step by Step
$\sqrt{2x-3}=3-x$
$(\sqrt{2x-3})^2=(3-x)^2$
$2x-3=9-6x+x^2$
$2x=12-6x+x^2$
$0=12-8x+x^2$
$x^2-8x+12=0$
$(x-2)(x-6)=0$
$x-2=0$
$x=2$
$x-6=0$
$x=6$
$\sqrt{2x-3}=3-x$
$\sqrt{2*2-3}=3-2$
$\sqrt{4-3}=1$
$\sqrt{1}=1$
$1=1$ (true)
$\sqrt{2x-3}=3-x$
$\sqrt{2*6-3}=3-6$
$\sqrt{12-3}=-3$
$\sqrt{9}=-3$
$3\ne-3$ (false)
