Question

how do you solve 2=\sqrt{3b-2}-\sqrt{10-b}\]

Answer 1

\[2=\sqrt{3b-2}-\sqrt{10-b}\quad,\quad b\in[2/3,10]\]
\[4=3b-2+10-b-2\sqrt{(3b-2)(10-b)}\]
\[2\sqrt{(3b-2)(10-b)}=2b+4\]
\[\sqrt{-3b^2+32b-20}=b+2\]
\[-3b^2+32b-20=b^2+4b+4\]
\[0=4b^2-28b+24=4(b-1)(b-6)\]
\[b=6\]