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

1 Answer
Feb 14, 2016

Answer:

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]}