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

1 Answer
Jan 6, 2016

Answer:

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

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

Explanation:

#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#