HELPPP

$\sqrt{7+\sqrt{4-\sqrt{3+x}}}=3$

$7+\sqrt{4-\sqrt{3+x}}=9$     (squared both sides)

$\sqrt{4-\sqrt{3+x}}=2$             (subtracted 7 from both sides)

$4-\sqrt{3+x}=4$                 (square both sides again)

$-\sqrt{3+x}=0$                     (subtracted 4 from both sides)

$\sqrt{3+x}=0$                        (divided both sides by -1)

$3+x=0$                            (squared both sides again)

$\boxed{x=-3}$