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

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

Find the zeros of y^2-4y+1=0

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

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

y = 2 + 3, y = 2 - 3

Decimal Notation

y = 3.73205 \dots, y = 0.26794 \dots

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = 2 + 3, y = 2 - 3

x^{2} + 20 x + 109 = 0

x^{2} - 8 x - 105 = 0

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

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

x^{2} - \frac{1}{2} x

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