what are the solutions to the equation

z^2 - 6z - 27 = 0 there are two solutions, and the options available are 3, -3, 9, and -9

well, geez, did you try them?

Or, note that you want (z-9)(z+3) = 0

oh yeah me answer was 3, -9, but that doesn't feel right, or it may just be me

oh so its -9, 3?

oh wait, -9, 3 isnt an answer,um, is it -3, 9?

now im confused

yeah all of the answers start with either 3 or -3

To find the solutions to the equation z^2 - 6z - 27 = 0, we can use the quadratic formula. The quadratic formula states that for an equation in the form ax^2 + bx + c = 0, the solutions are given by:

x = (-b ± √(b^2 - 4ac)) / (2a)

Comparing this with our equation, we have a = 1, b = -6, and c = -27. Now we can substitute these values into the quadratic formula and solve for z:

z = (-(-6) ± √((-6)^2 - 4(1)(-27))) / (2(1))

Simplifying further:

z = (6 ± √(36 + 108)) / 2
z = (6 ± √(144)) / 2
z = (6 ± 12) / 2

This gives us two possible solutions:

z1 = (6 + 12) / 2 = 18 / 2 = 9
z2 = (6 - 12) / 2 = -6 / 2 = -3

Therefore, the solutions to the equation z^2 - 6z - 27 = 0 are z = 9 and z = -3.

Given the options of 3, -3, 9, and -9, the correct solutions are 9 and -3.

good intuition. You got it exactly backwards.