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

1 Answer
Meave60
Feb 14, 2016

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]}

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