How do you find the zeros of y = 7x^2 + 5x – 3 using the quadratic formula?

1 Answer
Feb 4, 2016

x=(-5+sqrt(109))/14

x=(-5-sqrt(109))/14

Explanation:

y=7x^2+5x-3 is a quadratic equation in standard form, x^2+bx+c, where a=7, b=5, c=-3.

The zeroes of a quadratic equation are the x-intercepts where y=0. To determine the zeroes, solve for x.

Quadratic Formula

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

Substitute the known values from the quadratic equation into the quadratic formula.

x=(-5+-sqrt(5^2-(4*7*-3)))/(2*7)

Simplify.

x=(-5+-sqrt(25-(-84)))/14

Simplify.

x=(-5+-sqrt(109))/14

109 is a prime number.

Solve for x.

x=(-5+sqrt(109))/14

x=(-5-sqrt(109))/14