# How do you solve w^2/24-w/2+13/6=0 by completing the square?

Jul 18, 2016

The solutions will be $w = 6 \pm 4 i$.

#### Explanation:

We can start by removing fractions from the mix by multiplying both sides by $24$:
${w}^{2} - 12 w + 52 = 0$

Now observing that we need an equation looking like $w + b$ where $2 b = - 12$ it is clear that the squared term will be $w - 6$.

Since ${\left(w - 6\right)}^{2} = {w}^{2} - 12 w + 36$ we can take $36$ out of $52$, this gives us:
${\left(w - 6\right)}^{2} + 16 = 0$

we can manipulate this:
${\left(w - 6\right)}^{2} = - 16$

And take the square root of both sides:
$w - 6 = \pm 4 i$
$w = 6 \pm 4 i$

You can check this answer by inputting the coefficients into the quadratic equation as well.