What are the solutions to the equation 5y^2-7y+1=0 ?

The solutions to the equation 5y^2-7y+1=0 are y=(7+sqrt(29))/(10),y=(7-sqrt(29))/(10)

Find the zeros of 5y^2-7y+1=0

The zeros of 5y^2-7y+1=0 are y=(7+sqrt(29))/(10),y=(7-sqrt(29))/(10)

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5 y^{2} - 7 y + 1 = 0

y = \frac{7 + 29}{10}, y = \frac{7 - 29}{10}

Decimal Notation

y = 1.23851 \dots, y = 0.16148 \dots

Solution steps

5 y^{2} - 7 y + 1 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 5 \cdot 1}{2 \cdot 5}

(- 7)^{2} - 4 \cdot 5 \cdot 1 = 29

y_{1, 2} = \frac{- ( - 7 ) \pm 29}{2 \cdot 5}

Separate the solutions

y_{1} = \frac{- ( - 7 ) + 29}{2 \cdot 5}, y_{2} = \frac{- ( - 7 ) - 29}{2 \cdot 5}

y = \frac{- ( - 7 ) + 29}{2 \cdot 5} : \frac{7 + 29}{10}

y = \frac{- ( - 7 ) - 29}{2 \cdot 5} : \frac{7 - 29}{10}

The solutions to the quadratic equation are:

y = \frac{7 + 29}{10}, y = \frac{7 - 29}{10}

3 x (x - 2) + 5 (x - 2)

8 x^{6} + 729

x (x + 3) (x - 4) \leq 0

\frac{( n + 1 ) !}{n !}

2 = \frac{x - 21.4}{5.1}

What are the solutions to the equation 5y^2-7y+1=0 ?
The solutions to the equation 5y^2-7y+1=0 are y=(7+sqrt(29))/(10),y=(7-sqrt(29))/(10)
Find the zeros of 5y^2-7y+1=0
The zeros of 5y^2-7y+1=0 are y=(7+sqrt(29))/(10),y=(7-sqrt(29))/(10)