What are the solutions to the equation 3y^2-8y+1=0 ?

The solutions to the equation 3y^2-8y+1=0 are y=(4+sqrt(13))/3 ,y=(4-sqrt(13))/3

Find the zeros of 3y^2-8y+1=0

The zeros of 3y^2-8y+1=0 are y=(4+sqrt(13))/3 ,y=(4-sqrt(13))/3

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3 y^{2} - 8 y + 1 = 0

y = \frac{4 + 13}{3}, y = \frac{4 - 13}{3}

Decimal Notation

y = 2.53518 \dots, y = 0.13148 \dots

Solution steps

3 y^{2} - 8 y + 1 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 3 \cdot 1}{2 \cdot 3}

(- 8)^{2} - 4 \cdot 3 \cdot 1 = 213

y_{1, 2} = \frac{- ( - 8 ) \pm 2 13}{2 \cdot 3}

Separate the solutions

y_{1} = \frac{- ( - 8 ) + 2 13}{2 \cdot 3}, y_{2} = \frac{- ( - 8 ) - 2 13}{2 \cdot 3}

y = \frac{- ( - 8 ) + 2 13}{2 \cdot 3} : \frac{4 + 13}{3}

y = \frac{- ( - 8 ) - 2 13}{2 \cdot 3} : \frac{4 - 13}{3}

The solutions to the quadratic equation are:

y = \frac{4 + 13}{3}, y = \frac{4 - 13}{3}

18 x^{2} + 48 x = - 32

m, 2 (m^{2} + n^{2}) = 58

- \frac{1}{4} b^{2} - b^{2} = - 1

6 x - 8 = 2 x^{2}

6 x - 12 x^{2} = 0

What are the solutions to the equation 3y^2-8y+1=0 ?
The solutions to the equation 3y^2-8y+1=0 are y=(4+sqrt(13))/3 ,y=(4-sqrt(13))/3
Find the zeros of 3y^2-8y+1=0
The zeros of 3y^2-8y+1=0 are y=(4+sqrt(13))/3 ,y=(4-sqrt(13))/3