What is the answer to 16y(2-y)+(4y-5)^2=0 ?

The answer to 16y(2-y)+(4y-5)^2=0 is y= 25/8

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16y(2-y)+(4y-5)^2=0

16 y (2 - y) + (4 y - 5)^{2} = 0

y = \frac{25}{8}

Solution steps

16 y (2 - y) + (4 y - 5)^{2} = 0

Expand

16 y (2 - y) + (4 y - 5)^{2} : - 8 y + 25

- 8 y + 25 = 0

Move

25

to the right side

- 8 y = - 25

Divide both sides by

- 8

y = \frac{25}{8}

(x - 1) (x + 2) (x - 3) > 0

- (- x y^{2} z^{3})^{5} (x^{4} yz)^{2}

x^{2} = 4 x - 9

10 x^{5} y + 2 x^{3} y^{2} + 14 x^{2} y^{4}

i^{- 3}

What is the answer to 16y(2-y)+(4y-5)^2=0 ?
The answer to 16y(2-y)+(4y-5)^2=0 is y= 25/8
Solve for y: 16y(2-y)+(4y-5)^2=0
The solution is y= 25/8