Question

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

Answer 1

Zeros

`x=(-3+sqrt 89)/10` `~~0.6434`

`x=(-3-sqrt 89)/10` `~~`-1.2434#

Explanation:

Substitute 0 for `y`.

`5x^2+3x-4=0` is a quadratic equation in standard form, `ax^2+bx+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.

Quadratic Formula

`x=(-b+-sqrt(b^2-4ac))/(2a)`

`x=(-3+-sqrt(-3^2-(4*5*-4)))/(2*5)`

Simplify.

`x=(-3+-sqrt(9-(-80)))/10`

Simplify.

`x=(-3+-sqrt89)/10`

Zeros

`x=(-3+sqrt 89)/10` `~~0.6434`

`x=(-3-sqrt 89)/10` `~~`,`-1.2434`

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