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

1 Answer
George C.
Dec 18, 2015

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

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