How do you find the zeros, real and imaginary, of # y = 5x^2-30x+49 # using the quadratic formula?
1 Answer
Dec 18, 2015
See explanation...
Explanation:
This has zeros given by the quadratic formula:
#x = (-b+-sqrt(b^2-4ac))/(2a)#
#=(30+-sqrt((-30)^2-(4*5*49)))/(2*5)#
#=(30+-sqrt(900-980))/10#
#=(30+-sqrt(-80))/10#
#=(30+-sqrt(4^2*5) i)/10#
#=(30+-4sqrt(5)i)/10#
#=(15+-2sqrt(5)i)/5#
#=3+-(2sqrt(5))/5 i#
