What are the solutions to the equation z^2+2z-7=0 ?

The solutions to the equation z^2+2z-7=0 are z=-1+2sqrt(2),z=-1-2sqrt(2)

Find the zeros of z^2+2z-7=0

The zeros of z^2+2z-7=0 are z=-1+2sqrt(2),z=-1-2sqrt(2)

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

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

Decimal Notation

z = 1.82842 \dots, z = - 3.82842 \dots

Solution steps

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

9 x^{2} - 57 x - 20 = 22

x^{2} + 7 x = 144

s^{2} + 2 s + 1.49 = 0

49 b = b^{2} + 12 b + 36

n^{2} - 4 = - 4

What are the solutions to the equation z^2+2z-7=0 ?
The solutions to the equation z^2+2z-7=0 are z=-1+2sqrt(2),z=-1-2sqrt(2)
Find the zeros of z^2+2z-7=0
The zeros of z^2+2z-7=0 are z=-1+2sqrt(2),z=-1-2sqrt(2)