# How do you find the zeros, if any, of y= -3x^2 +1 3x+31 using the quadratic formula?

Aug 25, 2016

Zeros of $y \left(x\right) = - 3 {x}^{2} + 13 x + 31$ are $\frac{13}{6} + \frac{\sqrt{541}}{6}$ and $\frac{13}{6} - \frac{\sqrt{541}}{6}$.

#### Explanation:

According to quadratic formula, zeros of $y \left(x\right) = a {x}^{2} + b x + c$ are given by $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$.

Hence, zeros of $y \left(x\right) = - 3 {x}^{2} + 13 x + 31$ are given by

x=(-13+-sqrt(13^2-4×(-3)×31))/(2×(-3)

= $\frac{- 13 \pm \sqrt{169 + 372}}{- 6}$

= $\frac{- 13 \pm \sqrt{541}}{- 6}$

= $\frac{13}{6} \pm \frac{\sqrt{541}}{6}$

Hence zeros of $y \left(x\right) = - 3 {x}^{2} + 13 x + 31$ are $\frac{13}{6} + \frac{\sqrt{541}}{6}$ and $\frac{13}{6} - \frac{\sqrt{541}}{6}$.