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

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Answer 1 · 2016-02-04T02:56:51

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

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

#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#.

#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#