What are the solutions to the equation z^2-18z+27=-9 ?

The solutions to the equation z^2-18z+27=-9 are z=3(3+sqrt(5)),z=3(3-sqrt(5))

Find the zeros of z^2-18z+27=-9

The zeros of z^2-18z+27=-9 are z=3(3+sqrt(5)),z=3(3-sqrt(5))

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z^{2} - 18 z + 27 = - 9

z = 3 (3 + 5), z = 3 (3 - 5)

Decimal Notation

z = 15.70820 \dots, z = 2.29179 \dots

Solution steps

z^{2} - 18 z + 27 = - 9

Move

9

to the left side

z^{2} - 18 z + 36 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 1 \cdot 36}{2 \cdot 1}

(- 18)^{2} - 4 \cdot 1 \cdot 36 = 65

z_{1, 2} = \frac{- ( - 18 ) \pm 6 5}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- ( - 18 ) + 6 5}{2 \cdot 1}, z_{2} = \frac{- ( - 18 ) - 6 5}{2 \cdot 1}

z = \frac{- ( - 18 ) + 6 5}{2 \cdot 1} : 3 (3 + 5)

z = \frac{- ( - 18 ) - 6 5}{2 \cdot 1} : 3 (3 - 5)

The solutions to the quadratic equation are:

z = 3 (3 + 5), z = 3 (3 - 5)

4 x + 2 \leq - 10 and 2 x \leq 0

0 = x^{2} - 2 x + 3

\frac{x - 4 y}{5 x} - \frac{4 x}{3 x}

\frac{x}{2} \geq 10

6 x (x - 7) < 2 (3 x - 1) (x - 8)

What are the solutions to the equation z^2-18z+27=-9 ?
The solutions to the equation z^2-18z+27=-9 are z=3(3+sqrt(5)),z=3(3-sqrt(5))
Find the zeros of z^2-18z+27=-9
The zeros of z^2-18z+27=-9 are z=3(3+sqrt(5)),z=3(3-sqrt(5))