What are the solutions to the equation 9y^2+18y-12=6y ?

The solutions to the equation 9y^2+18y-12=6y are y= 2/3 ,y=-2

Find the zeros of 9y^2+18y-12=6y

The zeros of 9y^2+18y-12=6y are y= 2/3 ,y=-2

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9 y^{2} + 18 y - 12 = 6 y

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

Decimal Notation

y = 0.66666 \dots, y = - 2

Solution steps

9 y^{2} + 18 y - 12 = 6 y

Move

6 y

to the left side

9 y^{2} + 12 y - 12 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 9 ( - 12 )}{2 \cdot 9}

1 2^{2} - 4 \cdot 9 (- 12) = 24

y_{1, 2} = \frac{- 12 \pm 24}{2 \cdot 9}

Separate the solutions

y_{1} = \frac{- 12 + 24}{2 \cdot 9}, y_{2} = \frac{- 12 - 24}{2 \cdot 9}

y = \frac{- 12 + 24}{2 \cdot 9} : \frac{2}{3}

y = \frac{- 12 - 24}{2 \cdot 9} : - 2

The solutions to the quadratic equation are:

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

∣ x + 2 ∣ = 10

(\frac{2 x ^{4} y ^{2}}{4 x ^{2} y})^{3}

(6 x - 18)^{2} - 20 = - 25

y^{2} = 9

7 x + 15 = 3 (3 x - 7)

What are the solutions to the equation 9y^2+18y-12=6y ?
The solutions to the equation 9y^2+18y-12=6y are y= 2/3 ,y=-2
Find the zeros of 9y^2+18y-12=6y
The zeros of 9y^2+18y-12=6y are y= 2/3 ,y=-2