What are the solutions to the equation y^2-3y+1=0 ?

The solutions to the equation y^2-3y+1=0 are y=(3+sqrt(5))/2 ,y=(3-sqrt(5))/2

Find the zeros of y^2-3y+1=0

The zeros of y^2-3y+1=0 are y=(3+sqrt(5))/2 ,y=(3-sqrt(5))/2

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y^{2} - 3 y + 1 = 0

y = \frac{3 + 5}{2}, y = \frac{3 - 5}{2}

Decimal Notation

y = 2.61803 \dots, y = 0.38196 \dots

Solution steps

y^{2} - 3 y + 1 = 0

Solve with the quadratic formula

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

(- 3)^{2} - 4 \cdot 1 \cdot 1 = 5

y_{1, 2} = \frac{- ( - 3 ) \pm 5}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 3 ) + 5}{2 \cdot 1}, y_{2} = \frac{- ( - 3 ) - 5}{2 \cdot 1}

y = \frac{- ( - 3 ) + 5}{2 \cdot 1} : \frac{3 + 5}{2}

y = \frac{- ( - 3 ) - 5}{2 \cdot 1} : \frac{3 - 5}{2}

The solutions to the quadratic equation are:

y = \frac{3 + 5}{2}, y = \frac{3 - 5}{2}

x^{3} - x

4 \cdot 41

2 x^{2} - 10 x + m x - 5 m

\frac{( 3 x + 9 y )}{3}

2 X + 1 \leq - 3 (5 X - 8)

What are the solutions to the equation y^2-3y+1=0 ?
The solutions to the equation y^2-3y+1=0 are y=(3+sqrt(5))/2 ,y=(3-sqrt(5))/2
Find the zeros of y^2-3y+1=0
The zeros of y^2-3y+1=0 are y=(3+sqrt(5))/2 ,y=(3-sqrt(5))/2