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

The solutions to the equation 2y^2-y-7=0 are y=(1+sqrt(57))/4 ,y=(1-sqrt(57))/4

Find the zeros of 2y^2-y-7=0

The zeros of 2y^2-y-7=0 are y=(1+sqrt(57))/4 ,y=(1-sqrt(57))/4

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2 y^{2} - y - 7 = 0

y = \frac{1 + 57}{4}, y = \frac{1 - 57}{4}

Decimal Notation

y = 2.13745 \dots, y = - 1.63745 \dots

Solution steps

2 y^{2} - y - 7 = 0

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 2 (- 7) = 57

y_{1, 2} = \frac{- ( - 1 ) \pm 57}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- ( - 1 ) + 57}{2 \cdot 2}, y_{2} = \frac{- ( - 1 ) - 57}{2 \cdot 2}

y = \frac{- ( - 1 ) + 57}{2 \cdot 2} : \frac{1 + 57}{4}

y = \frac{- ( - 1 ) - 57}{2 \cdot 2} : \frac{1 - 57}{4}

The solutions to the quadratic equation are:

y = \frac{1 + 57}{4}, y = \frac{1 - 57}{4}

x, f (x + 1) = - x^{2} + x + 1

x, x^{2} - 3 x^{2} y = C

3 k^{2} - 96 k + 92 = 0

(2 x + 10) (3 x - 15) = 0

12 x^{2} - 2 = 8 x

What are the solutions to the equation 2y^2-y-7=0 ?
The solutions to the equation 2y^2-y-7=0 are y=(1+sqrt(57))/4 ,y=(1-sqrt(57))/4
Find the zeros of 2y^2-y-7=0
The zeros of 2y^2-y-7=0 are y=(1+sqrt(57))/4 ,y=(1-sqrt(57))/4