Answer
$(-5/3, 3/4)$
Work Step by Step
$12x^2+11x\leq 15$
$12x^2+11x-15\leq 15-15$
$12x^2+11x-15\leq 0$
$(4x-3)(3x+5) \leq 0$
$4x-3 \leq 0$
$4x-3+3 \leq 0+3$
$4x \leq 3$
$4x/4 \leq 3/4$
$x \leq 3/4$
$3x+5\leq0$
$3x+5-5 \leq 0-5$
$3x \leq -5$
$3x/3 \leq -5/3$
$x \leq -5/3$
We have three sections: $(-∞,-5/3)$, $(-5/3, 3/4)$, and $(3/4, ∞)$. We need to test one value for $x$ in each section to determine if the section would be a solution set. Since we have the $\leq$ sign, we exclude the end points and use parentheses instead of brackets.
Let $x=-3$, $x=0$, and $x=2$
$x=-3$
$12x^2+11x\leq 15$
$12(-3)^2+11*(-3)\leq 15$
$12*9-33 \leq 15$
$108-33 \leq 15$
$65 \leq 15$ (false)
$x=0$
$12x^2+11x\leq 15$
$12*0^2+11*0\leq 15$
$12*0+0 \leq 15$
$0+0 \leq 15$
$0 \leq 15$ (true)
$x=2$
$12x^2+11x\leq 15$
$12*2^2+11*2\leq 15$
$12*4 +22 \leq 15$
$48 +22 \leq 15$
$70 \leq 15$ (false)
