Question

How do you find the zeros, real and imaginary, of `y=5(x-3)^2-56` using the quadratic formula?

Answer 1

`x = 3 +- 3,35`

Explanation:

Develop `y = 5(x^2 - 6x + 9) - 56 = 5x^2 - 30x + 45 - 56`
`y = 5x^2 - 30x - 11 = 0`
`D = d^2 = b^2 - 4ac = 900 + 220 = 1120` --> `d = +- 33.47`.
There are 2 real roots:
`x = 30/10 +- 33.47/10 = 3 +- 3.35`
x1 = 6.35 and x2 = 0.35