Answer
$$(-∞,\frac{5-\sqrt{10}}{3})∪(\frac{5+\sqrt{10}}{3},+∞)$$
Work Step by Step
Let $3x^2 \gt 10x -5 $ be a function $f$.
Express the inequality in the form $f(x) \gt 0$. Begin by the rewriting the inequality so that $0$ is on the right side.
$3x^2 \gt 10x -5$
$3x^2 -10x +5\gt 10x -5 -10x +5$
$3x^2 -10x +5\gt 0 $
Find the $x$-intercepts by solving $3x^2 -10x +5$
Use the quadratic formula: $x = \frac{-b±\sqrt{b^2-4ac}}{2a}$ where $a=3$, $b=-10$, $c=5$
$x = \frac{-(-10)±\sqrt{-10^2-4⋅3⋅5}}{2⋅3}$
$x = \frac{5+\sqrt{10}}{3}$ or $\frac{5-\sqrt{10}}{3}$
These $x$-intercepts serve as boundary points that separate the number line into intervals.
The boundary points of this equation therefore divide the number line into three intervals:
$(-∞, \frac{5-\sqrt{10}}{3})(\frac{5-\sqrt{10}}{3},\frac{5+\sqrt{10}}{3})(\frac{5-\sqrt{10}}{3}, +∞)$
Choose one test value within each interval and evaluate $f$ at that number.
$(-∞, \frac{5-\sqrt{10}}{3})$
Test Value = $0$
$f=3x^2-10x+5$
$f=3(0)^2-10(0)-5$
$f=5$
Conclusion: $f (x) > 0$ for all $x$ in $(-∞,\frac{5-\sqrt{10}}{3})$.
$(\frac{5-\sqrt{10}}{3}, \frac{5+\sqrt{10}}{3})$
Test Value = $1$
$f=3x^2-10x+5$
$f=3(1)^2-10(1)+5$
$f=-2$
Conclusion: $f (x) < 0$ for all $x$ in $(\frac{5-\sqrt{10}}{3}, \frac{5+\sqrt{10}}{3})$.
$(\frac{5+\sqrt{10}}{3},+∞)$
Test Value = $3$
$f=3x^2-10x+5$
$f=3(3)^2-10(3)+5$
$f=2$
Conclusion: $f (x) > 0$ for all $x$ in $(\frac{5+\sqrt{10}}{3},+∞)$.
Write the solution set, selecting the interval or intervals that satisfy the given inequality. We are interested in solving $f (x) \gt 0$, where $f (x) = 3x^2-10x-5$. Based on the solution above, $f(x)\gt0$ for all $x$ in $(-∞,\frac{5-\sqrt{10}}{3})$ or $(\frac{5+\sqrt{10}}{3},+∞)$.
Thus, the solution set of the given inequality is:
$$(-∞,\frac{5-\sqrt{10}}{3})∪(\frac{5+\sqrt{10}}{3},+∞)$$