What are the solutions to the equation (-3+y)^2+y^2=5 ?

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

Find the zeros of (-3+y)^2+y^2=5

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

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

y = 2, y = 1

Solution steps

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

Expand

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

9 - 6 y + 2 y^{2} = 5

Move

5

to the left side

2 y^{2} - 6 y + 4 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = 2, y = 1

2 x^{2} + 24 x + 64

x^{2} + 6 x y + 3 x + 18 y

\frac{1 0 ^{2}}{1 0 ^{2} - 4} - \frac{1}{10 - 2}

y + 5 = 8

2 x^{2} - 20 x - 48

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