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

Decimal Notation

z = 3.73205 \dots, z = 0.26794 \dots

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, x^{2} + 4 x + 8 = - 2 y^{2}

2 a^{2} = 1

(x + 6) (x - 4) = 0

x (2 x + 1) = 3 x (- 2 x - 1) - 3 (x^{2} + x + 1)

- 5 d + 2 = - 6 d^{2}

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)