you're making this one way too hard when it is extremely simple to do ...
3x + 4 < 0
3x < -4
x < -4/3
now for the quadratic inequality ...
x<sup>2</sup> - 3x - 12 < 0
factor ...
(x - 4)(x + 3) < 0
at x = 4 and x = -3, the left side of the inequality equals 0 ... that means that every other value of x makes the left side either positive or negative.
the values -3 and 4 break up the x-values into 3 regions ...
check any value in each region to see if it makes the left side positive or negative ...
<---------|--------------|--------->
.............-3................4.............
for x < -3, try x = -4 ... (-4 - 4)(-4 + 3) > 0 , all values in this region make the inequality false
for -3 < x < 4, try x = 0 ... (0 - 4)(0 + 3) < 0 , all values in this region make the inequality true
for x > 4, try x = 5 ... (5 - 4)(5 + 3) > 0 , all values in this region make the inequality false.
so ... the solution set for the quadratic inequality is -3 < x < 4
now, my question is this ...
are you supposed to find the solution set that satisfies both inequalities?
if so, then determine where they overlap.