What are the solutions to the equation z^2+3z-7=z ?

The solutions to the equation z^2+3z-7=z are z=-1+2sqrt(2),z=-1-2sqrt(2)

Find the zeros of z^2+3z-7=z

The zeros of z^2+3z-7=z are z=-1+2sqrt(2),z=-1-2sqrt(2)

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z^{2} + 3 z - 7 = z

z = - 1 + 22, z = - 1 - 22

Solution steps

z^{2} + 3 z - 7 = z

Move

z

to the left side

z^{2} + 2 z - 7 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 7 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 7) = 42

z_{1, 2} = \frac{- 2 \pm 4 2}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- 2 + 4 2}{2 \cdot 1}, z_{2} = \frac{- 2 - 4 2}{2 \cdot 1}

z = \frac{- 2 + 4 2}{2 \cdot 1} : - 1 + 22

z = \frac{- 2 - 4 2}{2 \cdot 1} : - 1 - 22

The solutions to the quadratic equation are:

z = - 1 + 22, z = - 1 - 22

x^{2} + 2 x - 8 = 0

3 x^{2} - 7 = 5

7 k = 78 - k^{2}

12 x^{2} - 36 x = 0

x^{2} + x - 14 = 0

What are the solutions to the equation z^2+3z-7=z ?
The solutions to the equation z^2+3z-7=z are z=-1+2sqrt(2),z=-1-2sqrt(2)
Find the zeros of z^2+3z-7=z
The zeros of z^2+3z-7=z are z=-1+2sqrt(2),z=-1-2sqrt(2)