Answer
$x = \frac{-3 ± 3\sqrt {5}}{2}$
Work Step by Step
Let's rewrite this equation so that all terms are on the left side and $0$ is on the right side, meaning we rewrite the equation in standard form:
$3x^2 + 9x - 27 = 0$
Let's divide this equation by $3$ to simplify:
$x^2 + 3x - 9 = 0$
We are asked to solve this equation using the quadratic formula, which is given by: $x = \frac{-b ± \sqrt {b^2 - 4ac}}{2a}$ where $a$ is the coefficient of the first term, $b$ is the coefficient of the 1st degree term, and $c$ is the constant. In this exercise, $a = 1$, $b = 3$, and $c = -9$.
Let's plug in the values for $a$, $b$, and $c$ from our equation into the quadratic formula:
$x = \dfrac{-(3) ± \sqrt {(3)^2 - 4(1)(-9)}}{2(1)}$
Let's simplify:
$x = \dfrac{-3 ± \sqrt {9 + 36}}{2}$
Let's simplify what is inside the radical:
$x = \dfrac{-3 ± \sqrt {45}}{2}$
We can expand the square root of $45$ into its components: the square root of $9$ and the square root of $5$. In this way, we can take out the square root of $9$ (a perfect square) from under the radical and be left only with the square root of $5$:
$x = \dfrac{-3 ± 3\sqrt {5}}{2}$
