# How do you find the zeros, if any, of y= 3x^2 -2x+ 8 using the quadratic formula?

Feb 12, 2017

There are only complex zeros: $x = \frac{1}{3} \pm \frac{\sqrt{23}}{3} i$

#### Explanation:

Use the quadratic formula where $A = 3 , B = - 2 , C = 8$

$x = \left(- \left(- 2\right) \pm \frac{\sqrt{{\left(- 2\right)}^{2} - 4 \left(3\right) \left(8\right)}}{2 \left(3\right)}\right) = \left(\frac{2 \pm \sqrt{- 92}}{6}\right)$

$\sqrt{- 92} = \sqrt{4} \sqrt{- 23} = 2 \sqrt{23} i$

Therefore $x = \frac{1}{3} \pm \frac{\sqrt{23}}{3} i$

If you graph the function it does not cross the x-axis: graph{3x^2-2x+8 [-32.23, 31.8, -5.5, 26.55]}