# How do you find the zeros, real and imaginary, of  y = 5x^2-30x+49  using the quadratic formula?

Dec 18, 2015

See explanation...

#### Explanation:

$5 {x}^{2} - 30 x + 49$ is of the form $a {x}^{2} + b x + c$ with $a = 5$, $b = - 30$ and $c = 49$.

This has zeros given by the quadratic formula:

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

$= \frac{30 \pm \sqrt{{\left(- 30\right)}^{2} - \left(4 \cdot 5 \cdot 49\right)}}{2 \cdot 5}$

$= \frac{30 \pm \sqrt{900 - 980}}{10}$

$= \frac{30 \pm \sqrt{- 80}}{10}$

$= \frac{30 \pm \sqrt{{4}^{2} \cdot 5} i}{10}$

$= \frac{30 \pm 4 \sqrt{5} i}{10}$

$= \frac{15 \pm 2 \sqrt{5} i}{5}$

$= 3 \pm \frac{2 \sqrt{5}}{5} i$