How do you find the zeros, real and imaginary, of y=-5x^2-2x+29 using the quadratic formula?

1 Answer
Shwetank Mauria
Feb 22, 2016

Zeros of y=−5x^2−2x+29 are
-1/5+sqrt146/5 and -1/5+sqrt146/5

Explanation:

To find the zeros, real and imaginary, of y=−5x^2−2x+29, we need to identify values of x that make y=0 i.e. −5x^2−2x+29=0 or 5x^2+2x-29=0.

As solution of ax^2+bx+x=0 is given by (-b+-sqrt(b^2-4ac))/(2a) and as in 5x^2+2x-29=0, a=5, b=2, c=-29

Hence, zeros of 5x^2+2x-29=0 are given by

(-2+-sqrt(2^2-4*5*(-29)))/(2*5) or

(-2+-sqrt(4+580))/10 or (-2+-sqrt584)/10 or

(-2+-sqrt584)/10 or (-2+-2sqrt146)/10

Hence, zeros of y=−5x^2−2x+29 are
-1/5+sqrt146/5 and -1/5+sqrt146/5

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