Answer
$H$
Work Step by Step
We can solve this equation using the quadratic formula, which is given by:
$x = \dfrac{-b ± \sqrt {b^2 - 4ac}}{2a}$
where $a$ is the coefficient of the first term, $b$ is the coefficient of the 1st degree term, and $c$ is the constant.
Let's plug in the numbers from our equation into the formula:
$x = \dfrac{-6 ± \sqrt {6^2 - 4(3)(-5)}}{2(3)}$
Let's simplify:
$x = \dfrac{-6 ± \sqrt {36 + 60}}{6}$
Let's simplify what is inside the radical:
$x = \dfrac{-6 ± \sqrt {96}}{6}$
d
We can expand the square root of $96$ into its components: the square root of $16$ and the square root of $6$. In this way, we can take out the square root of $16$ (a perfect square) from under the radical and be left only with the square root of $6$:
$x = \dfrac{-6 ± 4\sqrt {6}}{6}$
Divide numerator and denominator by $2$ to simplify this fraction:
$x = \dfrac{-3 ± 2\sqrt {6}}{3}$
This solution matches option $H$.
