What are the solutions to the equation z(z+9)=1 ?

The solutions to the equation z(z+9)=1 are z=(-9+sqrt(85))/2 ,z=(-9-sqrt(85))/2

Find the zeros of z(z+9)=1

The zeros of z(z+9)=1 are z=(-9+sqrt(85))/2 ,z=(-9-sqrt(85))/2

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z (z + 9) = 1

z = \frac{- 9 + 85}{2}, z = \frac{- 9 - 85}{2}

Decimal Notation

z = 0.10977 \dots, z = - 9.10977 \dots

Solution steps

z (z + 9) = 1

Expand

z (z + 9) : z^{2} + 9 z

z^{2} + 9 z = 1

Move

1

to the left side

z^{2} + 9 z - 1 = 0

Solve with the quadratic formula

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

9^{2} - 4 \cdot 1 \cdot (- 1) = 85

z_{1, 2} = \frac{- 9 \pm 85}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- 9 + 85}{2 \cdot 1}, z_{2} = \frac{- 9 - 85}{2 \cdot 1}

z = \frac{- 9 + 85}{2 \cdot 1} : \frac{- 9 + 85}{2}

z = \frac{- 9 - 85}{2 \cdot 1} : \frac{- 9 - 85}{2}

The solutions to the quadratic equation are:

z = \frac{- 9 + 85}{2}, z = \frac{- 9 - 85}{2}

\frac{3}{5} + \frac{1}{6}

4 x + 1 = 9

\frac{1}{6} - \frac{7}{12}

9 m^{2} + 5 = - 7 m

(\frac{4}{9})^{- 2}

What are the solutions to the equation z(z+9)=1 ?
The solutions to the equation z(z+9)=1 are z=(-9+sqrt(85))/2 ,z=(-9-sqrt(85))/2
Find the zeros of z(z+9)=1
The zeros of z(z+9)=1 are z=(-9+sqrt(85))/2 ,z=(-9-sqrt(85))/2