What are the solutions to the equation (6+y)^2+y^2=218 ?

The solutions to the equation (6+y)^2+y^2=218 are y=7,y=-13

Find the zeros of (6+y)^2+y^2=218

The zeros of (6+y)^2+y^2=218 are y=7,y=-13

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(6 + y)^{2} + y^{2} = 218

y = 7, y = - 13

Solution steps

(6 + y)^{2} + y^{2} = 218

Expand

(6 + y)^{2} + y^{2} : 36 + 12 y + 2 y^{2}

36 + 12 y + 2 y^{2} = 218

Move

218

to the left side

2 y^{2} + 12 y - 182 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 2 ( - 182 )}{2 \cdot 2}

1 2^{2} - 4 \cdot 2 (- 182) = 40

y_{1, 2} = \frac{- 12 \pm 40}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- 12 + 40}{2 \cdot 2}, y_{2} = \frac{- 12 - 40}{2 \cdot 2}

y = \frac{- 12 + 40}{2 \cdot 2} : 7

y = \frac{- 12 - 40}{2 \cdot 2} : - 13

The solutions to the quadratic equation are:

y = 7, y = - 13

2 x^{2} - 4 x - 8 = - x^{2} + x

4 x^{2} + 9 = 73

2 x^{2} + 3 = - 7 x

x + x^{2} = 15

(2 x - 7) (x + 3) = 3

What are the solutions to the equation (6+y)^2+y^2=218 ?
The solutions to the equation (6+y)^2+y^2=218 are y=7,y=-13
Find the zeros of (6+y)^2+y^2=218
The zeros of (6+y)^2+y^2=218 are y=7,y=-13