Answer
The solutions are $x = \dfrac{2 - i\sqrt {2}}{2}, \dfrac{2 + i\sqrt {2}}{2}$.
Work Step by Step
We want to factor this polynomial first so we can find the solutions.
Rewrite the equation so that all terms are on the left side and it is equal to zero on the right side:
$2x^2 - 4x + 3 = 0$
The polynomial cannot be factored so we solve this equation using the quadratic 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 second term, and $c$ is the constant.
The equation above has $a=2, b=-4,$ and $c=3$. Substitute these into the formula to obtain:
$x = \dfrac{-(-4) \pm \sqrt {(-4)^2 - 4(2)(3)}}{2(2)}$
$x = \dfrac{4 \pm \sqrt {16 - 24}}{4}$
$x = \dfrac{4 \pm \sqrt {-8}}{4}$
$x = \dfrac{4 \pm \sqrt {(-4)(2)}}{4}$
We can take out $-4$ from the radical because the square root of $-4$ is $2i$:
$x = \dfrac{4 \pm 2i\sqrt {2}}{4}$
Divide all terms by $2$ to simplify this fraction:
$x = \dfrac{2 \pm i\sqrt {2}}{2}$
The solutions are $x = \dfrac{2 \pm i\sqrt {2}}{2}$.
