How do you find the zeros, real and imaginary, of $y= 8x^2+3x-2$ using the quadratic formula?

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Answer 1 · 2016-12-28T14:14:21

We have real zeros $(-3+sqrt73)/16$ and $(-3-sqrt73)/16$

Quadratic formula gives the zeros of a quadratic function $y=ax^2+bx+c$ as $(-b+-sqrt(b^2-4ac))/(2a)$ .

Hence, zeros of a function $y=8x^2+3x-2$ are

$(-3+-sqrt(3^2-4xx8xx(-2)))/(2xx8)$

or $(-3+-sqrt(9+64))/16$

or $(-3+-sqrt73)/16$

i.e. $(-3+sqrt73)/16$ and $(-3-sqrt73)/16$

i.e. we have real zeros.

How do you find the zeros, real and imaginary, of y= 8x^2+3x-2y= 8x^2+3x-2 using the quadratic formula?