Answer
The solutions are $x=\dfrac{5}{6}\pm\dfrac{\sqrt{145}}{6}$
Work Step by Step
$3x^{2}-5x-10=0$
Take out common factor $3$ from the left side of the equation:
$3\Big(x^{2}-\dfrac{5}{3}x-\dfrac{10}{3}\Big)=0$
Take the $3$ to divide the right side:
$x^{2}-\dfrac{5}{3}x-\dfrac{10}{3}=\dfrac{0}{3}$
$x^{2}-\dfrac{5}{3}x-\dfrac{10}{3}=0$
Take $\dfrac{10}{3}$ to the right side:
$x^{2}-\dfrac{5}{3}x=\dfrac{10}{3}$
Add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation. In this case, $b=-\dfrac{5}{3}$
$x^{2}-\dfrac{5}{3}x+\Big(-\dfrac{5}{3\cdot2}\Big)^{2}=\dfrac{10}{3}+\Big(-\dfrac{5}{3\cdot2}\Big)^{2}$
$x^{2}-\dfrac{5}{3}x+\dfrac{25}{36}=\dfrac{10}{3}+\dfrac{25}{36}$
$x^{2}-\dfrac{5}{3}x+\dfrac{25}{36}=\dfrac{145}{36}$
Factor the left side of the equation, which is a perfect square trinomial:
$\Big(x-\dfrac{5}{6}\Big)^{2}=\dfrac{145}{36}$
Take the square root of both sides:
$\sqrt{\Big(x-\dfrac{5}{6}\Big)^{2}}=\pm\sqrt{\dfrac{145}{36}}$
$x-\dfrac{5}{6}=\pm\dfrac{\sqrt{145}}{6}$
Solve for $x$:
$x=\dfrac{5}{6}\pm\dfrac{\sqrt{145}}{6}$
