What are the solutions to the equation 3(y^2-2)-y=0 ?

The solutions to the equation 3(y^2-2)-y=0 are y=(1+sqrt(73))/6 ,y=(1-sqrt(73))/6

Find the zeros of 3(y^2-2)-y=0

The zeros of 3(y^2-2)-y=0 are y=(1+sqrt(73))/6 ,y=(1-sqrt(73))/6

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3 (y^{2} - 2) - y = 0

y = \frac{1 + 73}{6}, y = \frac{1 - 73}{6}

Decimal Notation

y = 1.59066 \dots, y = - 1.25733 \dots

Solution steps

3 (y^{2} - 2) - y = 0

Expand

3 (y^{2} - 2) - y : 3 y^{2} - 6 - y

3 y^{2} - 6 - y = 0

Write in the standard form

a x^{2} + b x + c = 0

3 y^{2} - y - 6 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 3 ( - 6 )}{2 \cdot 3}

(- 1)^{2} - 4 \cdot 3 (- 6) = 73

y_{1, 2} = \frac{- ( - 1 ) \pm 73}{2 \cdot 3}

Separate the solutions

y_{1} = \frac{- ( - 1 ) + 73}{2 \cdot 3}, y_{2} = \frac{- ( - 1 ) - 73}{2 \cdot 3}

y = \frac{- ( - 1 ) + 73}{2 \cdot 3} : \frac{1 + 73}{6}

y = \frac{- ( - 1 ) - 73}{2 \cdot 3} : \frac{1 - 73}{6}

The solutions to the quadratic equation are:

y = \frac{1 + 73}{6}, y = \frac{1 - 73}{6}

4 x^{4} - 4 x^{2} y + y^{2}

- \frac{1}{4} \cdot \frac{3}{7}

2 x^{2} = 0

a^{2} - 11 a + 28

8 \cdot (- \frac{5}{2})

What are the solutions to the equation 3(y^2-2)-y=0 ?
The solutions to the equation 3(y^2-2)-y=0 are y=(1+sqrt(73))/6 ,y=(1-sqrt(73))/6
Find the zeros of 3(y^2-2)-y=0
The zeros of 3(y^2-2)-y=0 are y=(1+sqrt(73))/6 ,y=(1-sqrt(73))/6