What are the solutions to the equation 3y^2+6=11y ?

The solutions to the equation 3y^2+6=11y are y=3,y= 2/3

Find the zeros of 3y^2+6=11y

The zeros of 3y^2+6=11y are y=3,y= 2/3

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3 y^{2} + 6 = 11 y

y = 3, y = \frac{2}{3}

Decimal Notation

y = 3, y = 0.66666 \dots

Solution steps

3 y^{2} + 6 = 11 y

Move

11 y

to the left side

3 y^{2} + 6 - 11 y = 0

Write in the standard form

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

3 y^{2} - 11 y + 6 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 3 \cdot 6}{2 \cdot 3}

(- 11)^{2} - 4 \cdot 3 \cdot 6 = 7

y_{1, 2} = \frac{- ( - 11 ) \pm 7}{2 \cdot 3}

Separate the solutions

y_{1} = \frac{- ( - 11 ) + 7}{2 \cdot 3}, y_{2} = \frac{- ( - 11 ) - 7}{2 \cdot 3}

y = \frac{- ( - 11 ) + 7}{2 \cdot 3} : 3

y = \frac{- ( - 11 ) - 7}{2 \cdot 3} : \frac{2}{3}

The solutions to the quadratic equation are:

y = 3, y = \frac{2}{3}

y, 2 x^{3} - 7 x^{2} y + 2 y^{2} = q

x^{2} + 11 x + 16 = 0

x, 5 x^{2} - 8 x y + 5 y^{2} - 12 = 0

2 m^{2} - m - 10 = 0

- 2 x^{2} + 8 x - 7 = 0

What are the solutions to the equation 3y^2+6=11y ?
The solutions to the equation 3y^2+6=11y are y=3,y= 2/3
Find the zeros of 3y^2+6=11y
The zeros of 3y^2+6=11y are y=3,y= 2/3