How do you find the zeros, real and imaginary, of # y = 5x^2-30x+49 # using the quadratic formula?

1 Answer
Dec 18, 2015

Answer:

See explanation...

Explanation:

#5x^2-30x+49# is of the form #ax^2+bx+c# with #a=5#, #b=-30# and #c=49#.

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#