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

The solutions to the equation (3y^2)/2-y/4 = y/2 are y= 1/2 ,y=0

Find the zeros of (3y^2)/2-y/4 = y/2

The zeros of (3y^2)/2-y/4 = y/2 are y= 1/2 ,y=0

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\frac{3 y ^{2}}{2} - \frac{y}{4} = \frac{y}{2}

y = \frac{1}{2}, y = 0

Decimal Notation

y = 0.5, y = 0

Solution steps

\frac{3 y ^{2}}{2} - \frac{y}{4} = \frac{y}{2}

Find Least Common Multiplier of

2, 4 : 4

Multiply by LCM=

4

\frac{3 y ^{2}}{2} \cdot 4 - \frac{y}{4} \cdot 4 = \frac{y}{2} \cdot 4

Simplify

6 y^{2} - y = 2 y

Move

2 y

to the left side

6 y^{2} - 3 y = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 6 \cdot 0}{2 \cdot 6}

(- 3)^{2} - 4 \cdot 6 \cdot 0 = 3

y_{1, 2} = \frac{- ( - 3 ) \pm 3}{2 \cdot 6}

Separate the solutions

y_{1} = \frac{- ( - 3 ) + 3}{2 \cdot 6}, y_{2} = \frac{- ( - 3 ) - 3}{2 \cdot 6}

y = \frac{- ( - 3 ) + 3}{2 \cdot 6} : \frac{1}{2}

y = \frac{- ( - 3 ) - 3}{2 \cdot 6} : 0

The solutions to the quadratic equation are:

y = \frac{1}{2}, y = 0

2 x^{2} - 4 x y

- ln (\frac{1}{2})

(\frac{2 m ^{2}}{3 m ^{6} n ^{2}})^{4}

∣ 4 - 5 x ∣ \geq 6

10 x^{2} + 23 x - 21 = 0

What are the solutions to the equation (3y^2)/2-y/4 = y/2 ?
The solutions to the equation (3y^2)/2-y/4 = y/2 are y= 1/2 ,y=0
Find the zeros of (3y^2)/2-y/4 = y/2
The zeros of (3y^2)/2-y/4 = y/2 are y= 1/2 ,y=0