What are the solutions to the equation 2(-y+5)^2+y^2=20 ?

The solutions to the equation 2(-y+5)^2+y^2=20 are y=(10+sqrt(10))/3 ,y=(10-sqrt(10))/3

Find the zeros of 2(-y+5)^2+y^2=20

The zeros of 2(-y+5)^2+y^2=20 are y=(10+sqrt(10))/3 ,y=(10-sqrt(10))/3

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2 (- y + 5)^{2} + y^{2} = 20

y = \frac{10 + 10}{3}, y = \frac{10 - 10}{3}

Decimal Notation

y = 4.38742 \dots, y = 2.27924 \dots

Solution steps

2 (- y + 5)^{2} + y^{2} = 20

Expand

2 (- y + 5)^{2} + y^{2} : 3 y^{2} - 20 y + 50

3 y^{2} - 20 y + 50 = 20

Move

20

to the left side

3 y^{2} - 20 y + 30 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 3 \cdot 30}{2 \cdot 3}

(- 20)^{2} - 4 \cdot 3 \cdot 30 = 210

y_{1, 2} = \frac{- ( - 20 ) \pm 2 10}{2 \cdot 3}

Separate the solutions

y_{1} = \frac{- ( - 20 ) + 2 10}{2 \cdot 3}, y_{2} = \frac{- ( - 20 ) - 2 10}{2 \cdot 3}

y = \frac{- ( - 20 ) + 2 10}{2 \cdot 3} : \frac{10 + 10}{3}

y = \frac{- ( - 20 ) - 2 10}{2 \cdot 3} : \frac{10 - 10}{3}

The solutions to the quadratic equation are:

y = \frac{10 + 10}{3}, y = \frac{10 - 10}{3}

e^{2} = 73891

5 x^{2} + 7 x - j = 0

k (k - 5) = 0

(x - 1)^{2} - (2 x + 1) = 3 (x^{2} - 2)

x, 2 x^{2} + 32 y^{2} = - 4 x y

What are the solutions to the equation 2(-y+5)^2+y^2=20 ?
The solutions to the equation 2(-y+5)^2+y^2=20 are y=(10+sqrt(10))/3 ,y=(10-sqrt(10))/3
Find the zeros of 2(-y+5)^2+y^2=20
The zeros of 2(-y+5)^2+y^2=20 are y=(10+sqrt(10))/3 ,y=(10-sqrt(10))/3