# How do you find the zeros, real and imaginary, of y=3x^2+17x-43 using the quadratic formula?

Jul 18, 2016

We have real zeros $\frac{- 17 - \sqrt{805}}{6}$ or $\frac{- 17 + \sqrt{805}}{6}$

#### Explanation:

Zeros of $y = 3 {x}^{2} + 17 x - 43$ are those values of $x$ for which $y = 0$. Hence, zeros are given by the solution of quadratic equation $3 {x}^{2} + 17 x - 43 = 0$.

Quadratic formula gives solution of equation $a {x}^{2} + b x + c = 0$ as $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$. Hence for $3 {x}^{2} + 17 x - 43 = 0$, we have

$x = \frac{- 17 \pm \sqrt{{17}^{2} - 4 \cdot 3 \cdot \left(- 43\right)}}{2 \cdot 3}$

= $\frac{- 17 \pm \sqrt{289 + 516}}{6}$

= $\frac{- 17 \pm \sqrt{805}}{6}$

= $\frac{- 17 - \sqrt{805}}{6}$ or $\frac{- 17 + \sqrt{805}}{6}$,

Which are real zeros of $3 {x}^{2} + 17 x - 43 = 0$.