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

The solutions to the equation (-2y-1)^2+(y-1)^2=5 are y= 3/5 ,y=-1

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

The zeros of (-2y-1)^2+(y-1)^2=5 are y= 3/5 ,y=-1

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

y = \frac{3}{5}, y = - 1

Decimal Notation

y = 0.6, y = - 1

Solution steps

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

Expand

(- 2 y - 1)^{2} + (y - 1)^{2} : 5 y^{2} + 2 y + 2

5 y^{2} + 2 y + 2 = 5

Move

5

to the left side

5 y^{2} + 2 y - 3 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 5 ( - 3 )}{2 \cdot 5}

2^{2} - 4 \cdot 5 (- 3) = 8

y_{1, 2} = \frac{- 2 \pm 8}{2 \cdot 5}

Separate the solutions

y_{1} = \frac{- 2 + 8}{2 \cdot 5}, y_{2} = \frac{- 2 - 8}{2 \cdot 5}

y = \frac{- 2 + 8}{2 \cdot 5} : \frac{3}{5}

y = \frac{- 2 - 8}{2 \cdot 5} : - 1

The solutions to the quadratic equation are:

y = \frac{3}{5}, y = - 1

x^{2} - 16 x + 50 = 21

4 t^{2} - 19 t + 18 = 0

2 x^{2} + (2 x + 11)^{2} = x + 136

2 y^{2} - 7 y + 5 = 0

25 x^{2} - 4 x = 0

What are the solutions to the equation (-2y-1)^2+(y-1)^2=5 ?
The solutions to the equation (-2y-1)^2+(y-1)^2=5 are y= 3/5 ,y=-1
Find the zeros of (-2y-1)^2+(y-1)^2=5
The zeros of (-2y-1)^2+(y-1)^2=5 are y= 3/5 ,y=-1