What are the solutions to the equation ((23+2y)/5)^2-(23+2y)y+y^2=1 ?

The solutions to the equation ((23+2y)/5)^2-(23+2y)y+y^2=1 are y=-24,y=1

Find the zeros of ((23+2y)/5)^2-(23+2y)y+y^2=1

The zeros of ((23+2y)/5)^2-(23+2y)y+y^2=1 are y=-24,y=1

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(\frac{23 + 2 y}{5})^{2} - (23 + 2 y) y + y^{2} = 1

y = - 24, y = 1

Solution steps

(\frac{23 + 2 y}{5})^{2} - (23 + 2 y) y + y^{2} = 1

Expand

(\frac{23 + 2 y}{5})^{2} - (23 + 2 y) y + y^{2} : - \frac{21 y ^{2}}{25} - \frac{483 y}{25} + \frac{529}{25}

- \frac{21 y ^{2}}{25} - \frac{483 y}{25} + \frac{529}{25} = 1

Move

1

to the left side

\frac{- 21 y ^{2} - 483 y + 529}{25} - 1 = 0

Write in the standard form

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

- \frac{21 y ^{2}}{25} - \frac{483 y}{25} + \frac{529}{25} - 1 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - \frac{483}{25} ) \pm ( - \frac{483}{25} ) ^{2} - 4 ( - \frac{21}{25} ) ( \frac{529}{25} - 1 )}{2 ( - \frac{21}{25} )}

(- \frac{483}{25})^{2} - 4 (- \frac{21}{25}) (\frac{529}{25} - 1) = 21

y_{1, 2} = \frac{- ( - \frac{483}{25} ) \pm 21}{2 ( - \frac{21}{25} )}

Separate the solutions

y_{1} = \frac{- ( - \frac{483}{25} ) + 21}{2 ( - \frac{21}{25} )}, y_{2} = \frac{- ( - \frac{483}{25} ) - 21}{2 ( - \frac{21}{25} )}

y = \frac{- ( - \frac{483}{25} ) + 21}{2 ( - \frac{21}{25} )} : - 24

y = \frac{- ( - \frac{483}{25} ) - 21}{2 ( - \frac{21}{25} )} : 1

The solutions to the quadratic equation are:

y = - 24, y = 1

5 x^{2} + 8 x - 14 = 0

(m - 4)^{2} = 9

2 (x + 3)^{2} + 12 = 4

2 x^{2} - 7 x + 4 = 9

- x^{2} + 14 x - 45 = 0

What are the solutions to the equation ((23+2y)/5)^2-(23+2y)y+y^2=1 ?
The solutions to the equation ((23+2y)/5)^2-(23+2y)y+y^2=1 are y=-24,y=1
Find the zeros of ((23+2y)/5)^2-(23+2y)y+y^2=1
The zeros of ((23+2y)/5)^2-(23+2y)y+y^2=1 are y=-24,y=1