How do you find the zeros, real and imaginary, of #y=2x^2-5x+13 #using the quadratic formula?
1 Answer
Nov 22, 2015
Use the quadratic formula to find:
#x=5/4+-sqrt(79)/4 i#
Explanation:
This has zeros given by the quadratic formula:
#x=(-b+-sqrt(b^2-4ac))/(2a)#
#=(-(-5)+-sqrt((-5)^2-(4xx2xx13)))/(2*2)#
#=(5+-sqrt(25-104))/4#
#=5/4+-sqrt(-79)/4#
#=5/4+-sqrt(79)/4 i#
