Answer
$x: (-\frac{2}{3}, \frac{3}{2})$
Work Step by Step
$$6x^{2} - 6 \lt 5x$$
Re-writing the inequality results in $6x^{2} - 5x - 6 \lt0$ which is in a quadratic form. To find the values of x, we can use the quadratic formula $\frac{-b \frac{+}{} \sqrt {b^{2} - 4ac}}{2a}$:
$$\frac{-(-5) \frac{+}{} \sqrt {(-5)^{2} - 4(6)(-6)}}{2(6)}$$
$$\frac{5 \frac{+}{} \sqrt {25 + 144}}{12}$$
$$\frac{5 \frac{+}{} \sqrt {169}}{12}$$
$$\frac{5 \frac{+}{} 13}{12}$$
where $x$ is $\frac{5 + 13}{12} = \frac{18}{12} = \frac{3}{2}$, and also $\frac{5 - 13}{12} = \frac{-8}{12} = -\frac{2}{3}$
This means that we can write the original inequality in the following manner: $$(x-\frac{3}{2})(x+\frac{2}{3})\lt0$$ and solve each term independently:$$x-\frac{3}{2}\lt0$$$x\lt0+\frac{3}{2}$
$x\lt\frac{3}{2}$$$x + \frac{2}{3}\lt0$$$x\lt0-\frac{2}{3}$
$x\lt-\frac{2}{3}$
To find the domain of the inequality, we now use 3 values of $x$ (a value less than $-\frac{2}{3}$, a value between $-\frac{2}{3}$ and $\frac{3}{2}$, and a value greater than $\frac{3}{2}$) to test if they satisfy the conditions of the original inequality:$$6x^{2} - 6 \lt 5x$$$6(-1)^{2} - 6\lt5(-1)$
$0\lt-5$ which is not true.
$6(0)^{2} - 6\lt5(0)$
$-6\lt0$ which is true
$6(2)^{2} - 6\lt5(2)$
$6(4) - 6 \lt 10$
$18\lt10$ which is not true.
We therefore conclude that the domain for this inequality is {$x|-\frac{2}{3}\lt x \lt\frac{3}{2}$, where $x$ belongs to the realm of Real numbers}, also expressed as $x: (-\frac{2}{3}, \frac{3}{2})$ since neither term is included within the domain.
