Need help solving an equation by factoring.
PROBLEM: $3v^2-10v-12 = -28v + 36$
This is my solution, but it seems a little too much for school:
$3v^2+18v-48 = 0$
$3(v^2+18v-48) = 0$
$3((-1v + -8)(-1v + 2)) = 0$
$-1v + -8 = 0$
$v = -8$
$-1v + 2 = 0$
$v = 2$
Your algebra went astray: from $3v^2-10v-12=-28v+36$ you should get
$$3v^2+18v-48=0\;.$$
Then you can factor out the $3$ to get $3(v^2+6v-16)=0$, so $v^2+6v-16=0$, and this then factors nicely to give you $(v+8)(v-2)=0$.
You actually made compensating errors, since $3(v^2+18v-48)$ does not factor as
$$3\big((−1v+−8)(−1v+2)\big)\;;$$
your $$3\big((−1v+−8)(−1v+2)\big)=0$$ is actually equivalent to
$$3(v+8)(v-2)=0\;,$$
which is essentially what I got.