To find the solutions of the equation z^2 - 6z - 27 = 0, we can use the quadratic formula. The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions are given by:
x = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 1, b = -6, and c = -27. Substituting these values into the quadratic formula, we get:
z = (-(-6) ± √((-6)^2 - 4(1)(-27))) / (2(1))
= (6 ± √(36 + 108)) / 2
= (6 ± √144) / 2
= (6 ± 12) / 2
Thus, the solutions of the equation z^2 - 6z - 27 = 0 are:
z = (6 + 12) / 2 = 18 / 2 = 9
z = (6 - 12) / 2 = -6 / 2 = -3
Therefore, the solutions are z = 9 and z = -3.
What are the solutions of the equation?
z^2-6z-27=0.
1 answer