Algebra 1

No, Rosita is wrong. $x=-3$ is an extraneous solution.
$x=\sqrt {x+12}$ $-3=\sqrt {-3+12}$ $-3=\sqrt 9$ $-3\ne3$ (extraneous solution) $x=\sqrt {x+12}$ $4=\sqrt {4+12}$ $4=\sqrt {16}$ $4=4$