Using the quadratic formula, we can solve for z:
z = (-b ± sqrt(b^2 - 4ac)) / 2a
In this case, a = 1, b = -6, and c = -27. Substituting into the formula, we get:
z = (6 ± sqrt((-6)^2 - 4(1)(-27))) / 2(1)
z = (6 ± sqrt(36 + 108)) / 2
z = (6 ± sqrt(144)) / 2
z = (6 ± 12) / 2
This gives us two solutions:
z = 9 or z = -3
Therefore, the solutions to z^2 - 6z - 27 = 0 are z = 9 and z = -3.
z^2-6z-27=0
what are the solutions
1 answer