What are the solutions to the equation (3y^2)/2-2y=2 ?

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

Find the zeros of (3y^2)/2-2y=2

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

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

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

Decimal Notation

y = 2, y = - 0.66666 \dots

Solution steps

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

Multiply both sides by

2

3 y^{2} - 4 y = 4

Move

4

to the left side

3 y^{2} - 4 y - 4 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 3 (- 4) = 8

y_{1, 2} = \frac{- ( - 4 ) \pm 8}{2 \cdot 3}

Separate the solutions

y_{1} = \frac{- ( - 4 ) + 8}{2 \cdot 3}, y_{2} = \frac{- ( - 4 ) - 8}{2 \cdot 3}

y = \frac{- ( - 4 ) + 8}{2 \cdot 3} : 2

y = \frac{- ( - 4 ) - 8}{2 \cdot 3} : - \frac{2}{3}

The solutions to the quadratic equation are:

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

62 x^{2} + 60 x - 30 = 6770

y, x^{2} + (y - x)^{2} = 1

- x^{2} + 1.5 x + 5 = - 5

x^{2} + (- 7) x + 10 = 0

(2 + 3 t)^{2} + (1 - t)^{2} + (1 + 2 t)^{2} = 6

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