What are the solutions to the equation 3y^2-4y=2y+7y ?

The solutions to the equation 3y^2-4y=2y+7y are y= 13/3 ,y=0

Find the zeros of 3y^2-4y=2y+7y

The zeros of 3y^2-4y=2y+7y are y= 13/3 ,y=0

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3 y^{2} - 4 y = 2 y + 7 y

y = \frac{13}{3}, y = 0

Decimal Notation

y = 4.33333 \dots, y = 0

Solution steps

3 y^{2} - 4 y = 2 y + 7 y

Refine

3 y^{2} - 4 y = 9 y

Move

9 y

to the left side

3 y^{2} - 13 y = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 3 \cdot 0}{2 \cdot 3}

(- 13)^{2} - 4 \cdot 3 \cdot 0 = 13

y_{1, 2} = \frac{- ( - 13 ) \pm 13}{2 \cdot 3}

Separate the solutions

y_{1} = \frac{- ( - 13 ) + 13}{2 \cdot 3}, y_{2} = \frac{- ( - 13 ) - 13}{2 \cdot 3}

y = \frac{- ( - 13 ) + 13}{2 \cdot 3} : \frac{13}{3}

y = \frac{- ( - 13 ) - 13}{2 \cdot 3} : 0

The solutions to the quadratic equation are:

y = \frac{13}{3}, y = 0

- 33.2 x^{2} - 2.66 x + 0.04 = 0

y, 2 x^{2} + 3 y^{2} - 8 x + 6 y + 14 = 0

F, g b + N b = 2 F (b)^{2}

10 x^{2} = 16 x + 6 x^{2}

x^{2} - 6 x + 15 = - (x - 3)^{2} + 6

What are the solutions to the equation 3y^2-4y=2y+7y ?
The solutions to the equation 3y^2-4y=2y+7y are y= 13/3 ,y=0
Find the zeros of 3y^2-4y=2y+7y
The zeros of 3y^2-4y=2y+7y are y= 13/3 ,y=0