What are the solutions to the equation 2y^2+1=6y ?

The solutions to the equation 2y^2+1=6y are y=(3+sqrt(7))/2 ,y=(3-sqrt(7))/2

Find the zeros of 2y^2+1=6y

The zeros of 2y^2+1=6y are y=(3+sqrt(7))/2 ,y=(3-sqrt(7))/2

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2 y^{2} + 1 = 6 y

y = \frac{3 + 7}{2}, y = \frac{3 - 7}{2}

Decimal Notation

y = 2.82287 \dots, y = 0.17712 \dots

Solution steps

2 y^{2} + 1 = 6 y

Move

6 y

to the left side

2 y^{2} + 1 - 6 y = 0

Write in the standard form

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

2 y^{2} - 6 y + 1 = 0

Solve with the quadratic formula

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

(- 6)^{2} - 4 \cdot 2 \cdot 1 = 27

y_{1, 2} = \frac{- ( - 6 ) \pm 2 7}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- ( - 6 ) + 2 7}{2 \cdot 2}, y_{2} = \frac{- ( - 6 ) - 2 7}{2 \cdot 2}

y = \frac{- ( - 6 ) + 2 7}{2 \cdot 2} : \frac{3 + 7}{2}

y = \frac{- ( - 6 ) - 2 7}{2 \cdot 2} : \frac{3 - 7}{2}

The solutions to the quadratic equation are:

y = \frac{3 + 7}{2}, y = \frac{3 - 7}{2}

(\frac{2}{3})^{1}

\frac{x + 6}{x - 6} < 0

(3 x + 5) - 2 (- 7 x + 8) = 7 (- x + 2)

14 c^{3} - 42 c^{5} - 49 c^{4}

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What are the solutions to the equation 2y^2+1=6y ?
The solutions to the equation 2y^2+1=6y are y=(3+sqrt(7))/2 ,y=(3-sqrt(7))/2
Find the zeros of 2y^2+1=6y
The zeros of 2y^2+1=6y are y=(3+sqrt(7))/2 ,y=(3-sqrt(7))/2