# How do you find the zeros, real and imaginary, of y=3x^2+17x-3 using the quadratic formula?

Mar 12, 2016

The quadratic has 2 real roots at ${x}_{+} = \frac{5 \sqrt{13} - 17}{6} \cong 0.171$ and ${x}_{-} = \frac{- 17 - 5 \sqrt{13}}{6} \cong - 5.84$

#### Explanation:

To begin, we must put our quadratic in the standard form (which is it already) and set $y$ equal to zero.

$3 {x}^{2} + 17 x - 3 = 0$

The quadratic formula uses the form:

$a {x}^{2} + b x + c = 0$

where a,b, and c can be gotten from the equation above. The roots are given by:

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

If the roots are complex, we will get a negative number under the square root. In our case the roots are given by:

$x = \frac{- 17 \pm \sqrt{{17}^{2} - 4 \cdot 3 \cdot \left(- 3\right)}}{2 \cdot 3}$

$x = \frac{- 17 \pm 5 \sqrt{13}}{6}$

${x}_{+} = \frac{5 \sqrt{13} - 17}{6} \cong 0.171$

${x}_{-} = \frac{- 17 - 5 \sqrt{13}}{6} \cong - 5.84$