How do you find the roots, real and imaginary, of #y=3x^2+5x+2# using the quadratic formula?
1 Answer
Jan 12, 2016
Substitute the coefficients into the quadratic formula to find zeros:
#x=-1# and#x=-2/3#
Explanation:
It has zeros given by the quadratic formula:
#x=(-b+-sqrt(b^2-4ac))/(2a)#
#=(-5+-sqrt(5^2-(4*3*2)))/(2*3)#
#=(-5+-sqrt(25-24))/6#
#=(-5+-1)/6#
That is