How do you find the zeros, real and imaginary, of $y=3x^2-7x-5$ using the quadratic formula?

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Answer 1 · 2016-09-14T13:26:43

Zeros are real and are $(7-sqrt109)/6$ and $(7+sqrt109)/6$

The zeros of $y=ax^2+bx+c$ are given by quadratic formula

and are $(-b+-sqrt(b^2-4ac))/(2a)$

Hence, zeros of $y=3x^2-7x-5$ are given by

$(-(-7)+-sqrt((-7)^2-4xx3xx(-5)))/(2xx3)$

= $(7+-sqrt(49+60))/6$

= $(7+-sqrt109)/6$

Hence zeros are real and are $(7-sqrt109)/6$ and $(7+sqrt109)/6$

How do you find the zeros, real and imaginary, of y=3x^2-7x-5 y=3x^2-7x-5 using the quadratic formula?