Answer
$x=2$
Work Step by Step
$3x^2 > 2x + 5$ can be re-written as $3x^2 - 2x - 5 > 0$. To solve this exercise, one can initially treat the inequality as a quadratic equation and use the Quadratic Formula to solve for x:
$$x = \frac{-b\frac{+}{}\sqrt {b^{2} - 4ac}}{2a}$$
$$x = \frac{-(-2)\frac{+}{}\sqrt {(-2)^{2} - 4(3)(-5)}}{2(3)}$$
$$x = \frac{4\frac{+}{}\sqrt {4 + 60}}{6}$$
$$x = \frac{4\frac{+}{}\sqrt {64}}{6}$$
$$x = \frac{4\frac{+}{}8}{6}$$
where $x = \frac{4-8}{6} = \frac{-4}{6} = -\frac{2}{3}$ or $x = \frac{4+8}{6} = \frac{12}{6} = 2$
We can now use test values to identify the domain of the original inequality. $0$ is a good initial test value since it lies between $2$ and $-\frac{2}{3}$:
$$3(0)^2 - 2(0) - 5 > 0$$
$$-5>0$$
which is false, meaning that 0 lies outside the domain of the original inequality. Substituting $-2$ and $3$ gives us the following:
$$3(-2)^2 - 2(-2) - 5 > 0$$
$$12 +4 - 5 > 0$$
$$11>5$$
$$AND$$
$$3(3)^2 - 2(3) - 5 > 0$$
$$27 - 6 - 5>0$$
$$16>0$$
which both are TRUE. We can now accept that the domain of the original inequality is as follows: {$x|x=2$}
