Answer
$b=1$
Work Step by Step
$2b=\sqrt {b+3}$
$(2b)^2=(\sqrt {b+3})^2$
$4b^2=b+3$
$4b^2-b-3=0$
$4b^2+3b-4b-3=0$
$b(4b+3)-(4b+3)=0$
$(b-1)(4b+3)=0$
$b-1=0$ or $4b+3=0$
$b=1$ or $b=-\frac{3}{4}$
Check for extraneous solutions:
$2*1=\sqrt {1+3}=2$
$2*\frac{-3}{4}=\frac{-3}{2},\sqrt {\frac{-3}{4}+3}=3/2$ (eliminate this solution).
