# How do you find the zeros of  y = x^2 + 5x + 3  using the quadratic formula?

Nov 23, 2015

$\frac{- 5 \pm \sqrt{13}}{2}$

#### Explanation:

For the general quadratic equation $a {x}^{2} + b x + c$, the quadratic formula states that x=(-b+-sqrt(b^2-4ac))/(2a.

In your quadratic, $a = 1$, $b = 5$, and $c = 3$.

We can plug in values.

$x = \frac{- 5 \pm \sqrt{{5}^{2} - 4 \left(1\right) \left(3\right)}}{2 \left(1\right)} = \frac{- 5 \pm \sqrt{13}}{2}$