What are the solutions to the equation z^2+4z+1=0 ?

The solutions to the equation z^2+4z+1=0 are z=-2+sqrt(3),z=-2-sqrt(3)

Find the zeros of z^2+4z+1=0

The zeros of z^2+4z+1=0 are z=-2+sqrt(3),z=-2-sqrt(3)

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z^{2} + 4 z + 1 = 0

z = - 2 + 3, z = - 2 - 3

Solution steps

z^{2} + 4 z + 1 = 0

Solve with the quadratic formula

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

4^{2} - 4 \cdot 1 \cdot 1 = 23

z_{1, 2} = \frac{- 4 \pm 2 3}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- 4 + 2 3}{2 \cdot 1}, z_{2} = \frac{- 4 - 2 3}{2 \cdot 1}

z = \frac{- 4 + 2 3}{2 \cdot 1} : - 2 + 3

z = \frac{- 4 - 2 3}{2 \cdot 1} : - 2 - 3

The solutions to the quadratic equation are:

z = - 2 + 3, z = - 2 - 3

x^{2} + 7 x - 9 = 0

(x + 5) (2 x - 1) = 0

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

2 x^{2} - 6 x = - 1

k^{2} + 1 = 0

What are the solutions to the equation z^2+4z+1=0 ?
The solutions to the equation z^2+4z+1=0 are z=-2+sqrt(3),z=-2-sqrt(3)
Find the zeros of z^2+4z+1=0
The zeros of z^2+4z+1=0 are z=-2+sqrt(3),z=-2-sqrt(3)