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

The solutions to the equation 2y^2-y-8=0 are y=(1+sqrt(65))/4 ,y=(1-sqrt(65))/4

Find the zeros of 2y^2-y-8=0

The zeros of 2y^2-y-8=0 are y=(1+sqrt(65))/4 ,y=(1-sqrt(65))/4

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2 y^{2} - y - 8 = 0

y = \frac{1 + 65}{4}, y = \frac{1 - 65}{4}

Solution steps

2 y^{2} - y - 8 = 0

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 2 (- 8) = 65

y_{1, 2} = \frac{- ( - 1 ) \pm 65}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- ( - 1 ) + 65}{2 \cdot 2}, y_{2} = \frac{- ( - 1 ) - 65}{2 \cdot 2}

y = \frac{- ( - 1 ) + 65}{2 \cdot 2} : \frac{1 + 65}{4}

y = \frac{- ( - 1 ) - 65}{2 \cdot 2} : \frac{1 - 65}{4}

The solutions to the quadratic equation are:

y = \frac{1 + 65}{4}, y = \frac{1 - 65}{4}

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What are the solutions to the equation 2y^2-y-8=0 ?
The solutions to the equation 2y^2-y-8=0 are y=(1+sqrt(65))/4 ,y=(1-sqrt(65))/4
Find the zeros of 2y^2-y-8=0
The zeros of 2y^2-y-8=0 are y=(1+sqrt(65))/4 ,y=(1-sqrt(65))/4