Answer
$x = \dfrac{-5 \pm i\sqrt {47}}{4}$
Work Step by Step
Write the given equation in standard form to obtain
$$2x^2+5x+9=0$$
We are asked to use the quadratic equation to factor this polynomial. The quadratic equation is given by the formula:
$x = \dfrac{-b \pm \sqrt {b^2 - 4ac}}{2a}$
where $a$ is the coefficient of the $x^2$ term, $b$ is the coefficient of the $x$ term, and $c$ is the constant.
The given equation has $a=2, b=5, $ and $c=9$.
Plug these values into the formula to obtain:
$x = \dfrac{-5 \pm \sqrt {5^2 - 4(2)(9)}}{2(2)}$
$x = \dfrac{-5 \pm \sqrt {25 - 72}}{4}$
$x = \dfrac{-5 \pm \sqrt {-47}}{4}$
The number $-47$ can be factored as $47(-1)$ so:
$x = \dfrac{-5 \pm \sqrt {(47)(-1)}}{4}$
We can bring out the square root of $-1$ which is $i$, to obtain:
$x = \dfrac{-5 \pm i\sqrt {47}}{4}$
We cannot simplify this equation any further, so the solution is:
$x = \frac{-5 ± i\sqrt {47}}{4}$
