What are the solutions to the equation 4((9-4y)/2)^2+16y^2=20((9-4y)/2)+4y-19 ?

The solutions to the equation 4((9-4y)/2)^2+16y^2=20((9-4y)/2)+4y-19 are y= 5/8 ,y= 1/2

Find the zeros of 4((9-4y)/2)^2+16y^2=20((9-4y)/2)+4y-19

The zeros of 4((9-4y)/2)^2+16y^2=20((9-4y)/2)+4y-19 are y= 5/8 ,y= 1/2

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4 (\frac{9 - 4 y}{2})^{2} + 16 y^{2} = 20 (\frac{9 - 4 y}{2}) + 4 y - 19

y = \frac{5}{8}, y = \frac{1}{2}

Decimal Notation

y = 0.625, y = 0.5

Solution steps

4 (\frac{9 - 4 y}{2})^{2} + 16 y^{2} = 20 (\frac{9 - 4 y}{2}) + 4 y - 19

Expand

4 (\frac{9 - 4 y}{2})^{2} + 16 y^{2} : 81 - 72 y + 32 y^{2}

Expand

20 (\frac{9 - 4 y}{2}) + 4 y - 19 : - 36 y + 71

81 - 72 y + 32 y^{2} = - 36 y + 71

Move

71

to the left side

32 y^{2} - 72 y + 10 = - 36 y

Move

36 y

to the left side

32 y^{2} - 36 y + 10 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 36 ) \pm ( - 36 ) ^{2} - 4 \cdot 32 \cdot 10}{2 \cdot 32}

(- 36)^{2} - 4 \cdot 32 \cdot 10 = 4

y_{1, 2} = \frac{- ( - 36 ) \pm 4}{2 \cdot 32}

Separate the solutions

y_{1} = \frac{- ( - 36 ) + 4}{2 \cdot 32}, y_{2} = \frac{- ( - 36 ) - 4}{2 \cdot 32}

y = \frac{- ( - 36 ) + 4}{2 \cdot 32} : \frac{5}{8}

y = \frac{- ( - 36 ) - 4}{2 \cdot 32} : \frac{1}{2}

The solutions to the quadratic equation are:

y = \frac{5}{8}, y = \frac{1}{2}

5150

\frac{3 x - 7}{2} = \frac{x + 7}{3}

(\frac{2}{7})^{- 2}

x^{2} - 14 x = - 29

What are the solutions to the equation 4((9-4y)/2)^2+16y^2=20((9-4y)/2)+4y-19 ?
The solutions to the equation 4((9-4y)/2)^2+16y^2=20((9-4y)/2)+4y-19 are y= 5/8 ,y= 1/2
Find the zeros of 4((9-4y)/2)^2+16y^2=20((9-4y)/2)+4y-19
The zeros of 4((9-4y)/2)^2+16y^2=20((9-4y)/2)+4y-19 are y= 5/8 ,y= 1/2