How do you find the zeros, real and imaginary, of $y=3x^2+31x+9$ using the quadratic formula?

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Answer 1 · 2016-01-06T22:52:18

Identify an substitute the values of $a$ , $b$ and $c$ into the quadratic formula to find:

$x = (-31+-sqrt(853))/6$

$y=3x^2+31x+9$ is in the form $ax^2+bx+c$ with $a=3$ , $b=31$ and $c=9$ .

This has zeros given by the quadratic formula:

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

$=(-31+-sqrt(31^2-(4*3*9)))/(2*3)$

$=(-31+-sqrt(961-108))/6$

$=(-31+-sqrt(853))/6$

How do you find the zeros, real and imaginary, of y=3x^2+31x+9y=3x^2+31x+9 using the quadratic formula?