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

The solutions to the equation 4y=3y^2-4 are y=2,y=-2/3

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

The zeros of 4y=3y^2-4 are y=2,y=-2/3

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4 y = 3 y^{2} - 4

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

Decimal Notation

y = 2, y = - 0.66666 \dots

Solution steps

4 y = 3 y^{2} - 4

Switch sides

3 y^{2} - 4 = 4 y

Move

4 y

to the left side

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

Write in the standard form

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

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}

(m^{2} p^{5})^{3}

\frac{5}{- 15}

(\frac{1}{9})^{- 1}

16 - a^{2}

2 = \frac{1}{2} (5 x - 8)

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