# How do you find the zeros, real and imaginary, of y=5x^2+3x-4 using the quadratic formula?

Feb 14, 2016

Zeros

$x = \frac{- 3 + \sqrt{89}}{10}$$\approx 0.6434$

$x = \frac{- 3 - \sqrt{89}}{10}$$\approx$-1.2434#

#### Explanation:

Substitute 0 for $y$.

$5 {x}^{2} + 3 x - 4 = 0$ is a quadratic equation in standard form, $a {x}^{2} + b x + c$, where $a = 5 , b = 3 , c = - 4$.

The graph of a quadratic equation is a parabola. The zeros are the x-intercepts where the parabola crosses the y-axis, where $y = 0$. We find the zeros by using the quadratic formula.

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

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

Simplify.

$x = \frac{- 3 \pm \sqrt{9 - \left(- 80\right)}}{10}$

Simplify.

$x = \frac{- 3 \pm \sqrt{89}}{10}$

Zeros

$x = \frac{- 3 + \sqrt{89}}{10}$$\approx 0.6434$

$x = \frac{- 3 - \sqrt{89}}{10}$$\approx$,$- 1.2434$

graph{5x^2+3x-4 [-6.9, 7.29, -5.49, 1.605]}