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

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Answer 1 · 2016-05-18T07:09:44

Zeros of $y=-2x^2+x-3$ are

$-1/4+isqrt23/4$ and $-1/4-isqrt23/4$

If $y=ax^2+bx+c$ , quadratic formula states that zeros are given by

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

Hence, in $y=-2x^2+x-3$ , zeros are

$(-1+-sqrt(1^2-4*(-2)*(-3)))/(2*2)$

= $(-1+-sqrt(1-24))/(2*2)$

or $(-1+-sqrt(-23))/4=-1/4+-isqrt23/4$

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