What are the solutions to the equation 4((2y-3)/2)y-((2y-3)/2)^2-3((2y-3)/2)=0 ?

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

Find the zeros of 4((2y-3)/2)y-((2y-3)/2)^2-3((2y-3)/2)=0

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

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4 (\frac{2 y - 3}{2}) y - (\frac{2 y - 3}{2})^{2} - 3 (\frac{2 y - 3}{2}) = 0

y = \frac{3}{2}, y = \frac{1}{2}

Decimal Notation

y = 1.5, y = 0.5

Solution steps

4 (\frac{2 y - 3}{2}) y - (\frac{2 y - 3}{2})^{2} - 3 (\frac{2 y - 3}{2}) = 0

Expand

4 (\frac{2 y - 3}{2}) y - (\frac{2 y - 3}{2})^{2} - 3 (\frac{2 y - 3}{2}) : 3 y^{2} - 6 y + \frac{9}{4}

3 y^{2} - 6 y + \frac{9}{4} = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 3 \cdot \frac{9}{4}}{2 \cdot 3}

(- 6)^{2} - 4 \cdot 3 \cdot \frac{9}{4} = 3

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = \frac{3}{2}, y = \frac{1}{2}

14 y^{2} - 59 y + 12 = 0

2 x^{2} + 7 x = 1

0 = x^{2} + 2 x + 7

2 a^{2} = 25

z^{2} = 90

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