## anonymous 5 years ago how do you solve 2=\sqrt{3b-2}-\sqrt{10-b}\]

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