Answer
$x = \dfrac{3 \pm i\sqrt {23}}{4}$
Work Step by Step
We are asked to use the quadratic equation to factor this polynomial.
First, we want to rewrite the equation so that all the terms are on one side of the equation, and $0$ is on the other side of the equation:
$2(x^2+2)-3x=0\\
2x^2+4 - 3x = 0\\
2x^2-3x+4=0$
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 equation above has $a=2, b=-3, $ and $c=4$. Substitute these into the formula above to obtain:
$x = \dfrac{-(-3) \pm \sqrt {(-3)^2 - 4(2)(4)}}{2(2)}$
$x = \dfrac{3 \pm \sqrt {9 - 32}}{4}$
$x = \dfrac{3 \pm \sqrt {-23}}{4}$
The number $-23$ can be factored as $-1(23)$ so:
$x = \dfrac{3 \pm \sqrt {(-1)(23)}}{4}$
We bring out the square root of $-1$, which is $i$, to obtain:
$x = \dfrac{3 \pm i\sqrt {23}}{4}$
We cannot simplify this equation any further, so the solution is:
$x = \dfrac{3 \pm i\sqrt {23}}{4}$
