How do you find the zeros, real and imaginary, of $y= -3x^2-8x-18$ using the quadratic formula?

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Answer 1 · 2016-06-09T04:16:24

Zeros of $y=-3x^2-8x-18$ are

$-4/3-2sqrt38i$ and $-4/3+2sqrt38i$

Zeros of $y=ax^2+bx+c$ are given by the quadratic formula as

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

Hence for $y=-3x^2-8x-18$ these are given by

$(-(-8)+-sqrt((-8)^2-4xx(-3)xx(-18)))/(2*(-3))=(8+-sqrt(64-216))/(-6)$

= $-4/3+-sqrt(-152)/6=-4/3+-i2sqrt38$

Hence zeros of $y=-3x^2-8x-18$ are

$-4/3-2sqrt38i$ and $-4/3+2sqrt38i$

How do you find the zeros, real and imaginary, of y= -3x^2-8x-18 y= -3x^2-8x-18 using the quadratic formula?