What are the solutions to the equation z^2+2z-4=0 ?

The solutions to the equation z^2+2z-4=0 are z=-1+sqrt(5),z=-1-sqrt(5)

Find the zeros of z^2+2z-4=0

The zeros of z^2+2z-4=0 are z=-1+sqrt(5),z=-1-sqrt(5)

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z^{2} + 2 z - 4 = 0

z = - 1 + 5, z = - 1 - 5

Solution steps

z^{2} + 2 z - 4 = 0

Solve with the quadratic formula

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

2^{2} - 4 \cdot 1 \cdot (- 4) = 25

z_{1, 2} = \frac{- 2 \pm 2 5}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- 2 + 2 5}{2 \cdot 1}, z_{2} = \frac{- 2 - 2 5}{2 \cdot 1}

z = \frac{- 2 + 2 5}{2 \cdot 1} : - 1 + 5

z = \frac{- 2 - 2 5}{2 \cdot 1} : - 1 - 5

The solutions to the quadratic equation are:

z = - 1 + 5, z = - 1 - 5

5 c^{2} + 7 c + 1 = 2 c

q^{2} + 34 = 35

\frac{2}{10} - \frac{3}{10}

\frac{9 x ^{2} - 6 x + 1}{x ^{3} - x ^{2}} \leq 0

4 x^{2} + 10 x = - 3

What are the solutions to the equation z^2+2z-4=0 ?
The solutions to the equation z^2+2z-4=0 are z=-1+sqrt(5),z=-1-sqrt(5)
Find the zeros of z^2+2z-4=0
The zeros of z^2+2z-4=0 are z=-1+sqrt(5),z=-1-sqrt(5)