What are the solutions to the equation (3y+2)(y-1)=y(y+2) ?

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

Find the zeros of (3y+2)(y-1)=y(y+2)

The zeros of (3y+2)(y-1)=y(y+2) are y=2,y=-1/2

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(3 y + 2) (y - 1) = y (y + 2)

y = 2, y = - \frac{1}{2}

Decimal

y = 2, y = - 0.5

Solution steps

(3 y + 2) (y - 1) = y (y + 2)

Expand

(3 y + 2) (y - 1) : 3 y^{2} - y - 2

Expand

y (y + 2) : y^{2} + 2 y

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

Move

2 y

to the left side

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

Move

y^{2}

to the left side

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

Solve with the quadratic formula

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

(- 3)^{2} - 4 \cdot 2 (- 2) = 5

y_{1, 2} = \frac{- ( - 3 ) \pm 5}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- ( - 3 ) + 5}{2 \cdot 2}, y_{2} = \frac{- ( - 3 ) - 5}{2 \cdot 2}

y = \frac{- ( - 3 ) + 5}{2 \cdot 2} : 2

y = \frac{- ( - 3 ) - 5}{2 \cdot 2} : - \frac{1}{2}

The solutions to the quadratic equation are:

y = 2, y = - \frac{1}{2}

- 9 x + 4 x^{2} + 3 = 0

- 2 x^{2} + x + 9 = 0

2 x^{2} + 4 x + 11 = 0

a, n = 4 a^{2} - m

x^{2} - 5 x - 6

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