# How do you find the zeros of  y = 3x^2 - 13x +15  using the quadratic formula?

Aug 15, 2017

This sucker has complex roots!

#### Explanation:

$x = \frac{- B \pm \sqrt{\left({B}^{2} - 4 A C\right)}}{2 A}$

Plugging in A = 3, B = -13, C = 15 gives you:

$\frac{13 \pm \sqrt{169 - 180}}{6}$

or $\frac{13 \pm \sqrt{- 11}}{6}$

Which is your problem, since there is no square root of a negative number.

Best you can do is rewrite it as a COMPLEX number of form $a + b i$ where i is taken to be $\sqrt{- 1}$.

So you can do a little rewrite of $\sqrt{- 11}$ as $\sqrt{- 1 \cdot 11}$ = $\sqrt{11} \cdot i$

And then evaluate it once for the positive case of the $\pm$, and once for the negative. Which gives roots of:

$\frac{13}{6} + \left(\frac{\sqrt{11}}{6} \cdot i\right)$ = 2.1667 + 0.553i

And

$\frac{13}{6} - \left(\frac{\sqrt{11}}{6} \cdot i\right)$ = 2.1667 - 0.553i

(Note that I've rounded off.