# How do you find the zeros, real and imaginary, of y=2x^2-5x+13 using the quadratic formula?

Nov 22, 2015

Use the quadratic formula to find:

$x = \frac{5}{4} \pm \frac{\sqrt{79}}{4} i$

#### Explanation:

$2 {x}^{2} - 5 x + 13$ is in the form $a {x}^{2} + b x + c$ with $a = 2$, $b = - 5$ and $c = 13$.

This has zeros given by the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$= \frac{- \left(- 5\right) \pm \sqrt{{\left(- 5\right)}^{2} - \left(4 \times 2 \times 13\right)}}{2 \cdot 2}$

$= \frac{5 \pm \sqrt{25 - 104}}{4}$

$= \frac{5}{4} \pm \frac{\sqrt{- 79}}{4}$

$= \frac{5}{4} \pm \frac{\sqrt{79}}{4} i$