What are the solutions to the equation z^2+6z=1 ?

The solutions to the equation z^2+6z=1 are z=-3+sqrt(10),z=-3-sqrt(10)

Find the zeros of z^2+6z=1

The zeros of z^2+6z=1 are z=-3+sqrt(10),z=-3-sqrt(10)

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z^{2} + 6 z = 1

z = - 3 + 10, z = - 3 - 10

Decimal Notation

z = 0.16227 \dots, z = - 6.16227 \dots

Solution steps

z^{2} + 6 z = 1

Move

1

to the left side

z^{2} + 6 z - 1 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

6^{2} - 4 \cdot 1 \cdot (- 1) = 210

z_{1, 2} = \frac{- 6 \pm 2 10}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- 6 + 2 10}{2 \cdot 1}, z_{2} = \frac{- 6 - 2 10}{2 \cdot 1}

z = \frac{- 6 + 2 10}{2 \cdot 1} : - 3 + 10

z = \frac{- 6 - 2 10}{2 \cdot 1} : - 3 - 10

The solutions to the quadratic equation are:

z = - 3 + 10, z = - 3 - 10

3^{x} \geq 1

∣ 5 x - 9 ∣ + 9 = 4 x

2 x^{2} - x + 6 = - 4 x + 8

- 48 = 6 (x - 3)

\frac{1}{3} (48 n + 9) + 1 = \frac{1}{5} (30 + 30 n)

What are the solutions to the equation z^2+6z=1 ?
The solutions to the equation z^2+6z=1 are z=-3+sqrt(10),z=-3-sqrt(10)
Find the zeros of z^2+6z=1
The zeros of z^2+6z=1 are z=-3+sqrt(10),z=-3-sqrt(10)