Answer
$x = \dfrac{5 ± i \sqrt {11}}{6}$
Work Step by Step
We are asked to 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:
$x = \dfrac{-(-5) ± \sqrt {(-5)^2 - 4(3)(3)}}{2(3)}$
Let's simplify:
$x = \dfrac{5 ± \sqrt {25 - 36}}{6}$
Let's simplify what is inside the radical:
$x = \dfrac{5 ± \sqrt {-11}}{6}$
We can expand the square root of $-11$ into its components: the square root of $-1$ and the square root of $11$. In this way, we can take out the square root of $-1$ from under the radical and be left only with the square root of $11$:
$x = \dfrac{5 ± \sqrt {(-1)(11)}}{6}$
Take the square root of $-1$, which is $i$:
$x = \dfrac{5 ± i \sqrt {11}}{6}$
