Answer
$x = \dfrac{5 \pm \sqrt {5}}{2}$
Work Step by Step
Write the equation in standard form to obtain:
$$x^2-5x+5=0$$
We are asked to solve this equation using the quadratic formula, which is given by:
$x = \dfrac{-b \pm \sqrt {b^2 - 4ac}}{2a}$
where $a$ is the coefficient of the $x^2$ term, $b$ is the coefficient of the 1st degree term, and $c$ is the constant.
The equation above has $a=1, b=-5,$ and $c=5$. Substitute these numbers into the formula above to obtain:
$x = \dfrac{-(-5)\pm \sqrt {(-5)^2 - 4(1)(5)}}{2(1)}$
$x = \dfrac{5 \pm \sqrt {25 - 20}}{2}$
$x = \dfrac{5 \pm \sqrt {5}}{2}$
