Answer
$x=6$
Work Step by Step
Square both sides and put all terms on one side to create a quadratic equation. Then use trial and error to factor it, and set the values found equal to zero. One of the solutions does not work.
$\sqrt {x+3}=x-3$
$(\sqrt {x+3})^2=(x-3)^2$
$x+3=x^2-6x+9$
$x^2-7x+6=0$
$(x-1)(x-6)=0$
$x=1,6$
$\sqrt {1+3}\ne1-3$
$x=6$
