What are the solutions to the equation z^2-3z-9=0 ?

The solutions to the equation z^2-3z-9=0 are z=(3+3sqrt(5))/2 ,z=(3-3sqrt(5))/2

Find the zeros of z^2-3z-9=0

The zeros of z^2-3z-9=0 are z=(3+3sqrt(5))/2 ,z=(3-3sqrt(5))/2

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z^{2} - 3 z - 9 = 0

z = \frac{3 + 3 5}{2}, z = \frac{3 - 3 5}{2}

Decimal Notation

z = 4.85410 \dots, z = - 1.85410 \dots

Solution steps

z^{2} - 3 z - 9 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 9 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 9) = 35

z_{1, 2} = \frac{- ( - 3 ) \pm 3 5}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- ( - 3 ) + 3 5}{2 \cdot 1}, z_{2} = \frac{- ( - 3 ) - 3 5}{2 \cdot 1}

z = \frac{- ( - 3 ) + 3 5}{2 \cdot 1} : \frac{3 + 3 5}{2}

z = \frac{- ( - 3 ) - 3 5}{2 \cdot 1} : \frac{3 - 3 5}{2}

The solutions to the quadratic equation are:

z = \frac{3 + 3 5}{2}, z = \frac{3 - 3 5}{2}

7 + 3 a = 12 + 2 a

∣ x + 10 ∣ \leq 5

[6 - 2 w] < 10

\frac{4 x + 8}{x - 1} < 3

0.2 \cdot 0.4

What are the solutions to the equation z^2-3z-9=0 ?
The solutions to the equation z^2-3z-9=0 are z=(3+3sqrt(5))/2 ,z=(3-3sqrt(5))/2
Find the zeros of z^2-3z-9=0
The zeros of z^2-3z-9=0 are z=(3+3sqrt(5))/2 ,z=(3-3sqrt(5))/2