# How do you find the zeros, real and imaginary, of y= x^2-5x+16 using the quadratic formula?

Feb 3, 2016

$x = \frac{5 + i \sqrt{39}}{2}$ and $x = \frac{5 - i \sqrt{39}}{2}$

#### Explanation:

The quadratic formula is: $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$.

This is used to solve equations of the form $y = a {x}^{2} + b x + c$, which is the same type of equation as $y = {x}^{2} - 5 x + 16$. We can see that for our example, $a = 1$, $b = - 5$, $c = 16$. Plugging these guys into the formula,
$\frac{- \left(- 5\right) \pm \sqrt{{\left(- 5\right)}^{2} - 4 \left(1\right) \left(16\right)}}{2 \left(1\right)}$

And simplifying,
$\frac{5 \pm \sqrt{25 - 64}}{2}$

$\frac{5 \pm \sqrt{- 39}}{2}$

We now see we have an issue - a negative in the square root. That means both of our solutions will be imaginary, so to represent that, we take the $\sqrt{- 1}$ ($i$) out of the square root to make it positive:
$\frac{5 \pm i \sqrt{39}}{2}$

The next thing we look for is any way to simplify the square root. But $\sqrt{39}$ can't be simplified further, which means our solutions are:
$x = \frac{5 + i \sqrt{39}}{2}$
$x = \frac{5 - i \sqrt{39}}{2}$