What are the solutions to the equation (x^2-4x+1)=0 ?

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

Find the zeros of (x^2-4x+1)=0

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

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(x^{2} - 4 x + 1) = 0

x = 2 + 3, x = 2 - 3

Decimal Notation

x = 3.73205 \dots, x = 0.26794 \dots

Solution steps

(x^{2} - 4 x + 1) = 0

Refine

x^{2} - 4 x + 1 = 0

Solve with the quadratic formula

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

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

x_{1, 2} = \frac{- ( - 4 ) \pm 2 3}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 3}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 2 3}{2 \cdot 1}

x = \frac{- ( - 4 ) + 2 3}{2 \cdot 1} : 2 + 3

x = \frac{- ( - 4 ) - 2 3}{2 \cdot 1} : 2 - 3

The solutions to the quadratic equation are:

x = 2 + 3, x = 2 - 3

6 x^{2} = 8 x

x, z = - 3 x^{2} + y^{2} + 6 x - 8 y

x, s x (x) - x (0^{-}) 3 x (s) = 4 \frac{1}{( s ^{1} )}

(3 x - 2)^{2} = \frac{1}{4}

1 3^{2} = 1 2^{2} + b^{2}

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