# How do you find the zeros, real and imaginary, of y=-5x^2-2x+29 using the quadratic formula?

Feb 22, 2016

Zeros of y=−5x^2−2x+29 are
$- \frac{1}{5} + \frac{\sqrt{146}}{5}$ and $- \frac{1}{5} + \frac{\sqrt{146}}{5}$

#### Explanation:

To find the zeros, real and imaginary, of y=−5x^2−2x+29, we need to identify values of $x$ that make $y = 0$ i.e. −5x^2−2x+29=0 or $5 {x}^{2} + 2 x - 29 = 0$.

As solution of $a {x}^{2} + b x + x = 0$ is given by $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ and as in $5 {x}^{2} + 2 x - 29 = 0$, $a = 5 , b = 2 , c = - 29$

Hence, zeros of $5 {x}^{2} + 2 x - 29 = 0$ are given by

$\frac{- 2 \pm \sqrt{{2}^{2} - 4 \cdot 5 \cdot \left(- 29\right)}}{2 \cdot 5}$ or

$\frac{- 2 \pm \sqrt{4 + 580}}{10}$ or $\frac{- 2 \pm \sqrt{584}}{10}$ or

$\frac{- 2 \pm \sqrt{584}}{10}$ or $\frac{- 2 \pm 2 \sqrt{146}}{10}$

Hence, zeros of y=−5x^2−2x+29 are
$- \frac{1}{5} + \frac{\sqrt{146}}{5}$ and $- \frac{1}{5} + \frac{\sqrt{146}}{5}$