What are the solutions to the equation (9+y)^2+y^2=305 ?

The solutions to the equation (9+y)^2+y^2=305 are y=7,y=-16

Find the zeros of (9+y)^2+y^2=305

The zeros of (9+y)^2+y^2=305 are y=7,y=-16

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(9 + y)^{2} + y^{2} = 305

y = 7, y = - 16

Solution steps

(9 + y)^{2} + y^{2} = 305

Expand

(9 + y)^{2} + y^{2} : 81 + 18 y + 2 y^{2}

81 + 18 y + 2 y^{2} = 305

Move

305

to the left side

2 y^{2} + 18 y - 224 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 \cdot 2 ( - 224 )}{2 \cdot 2}

1 8^{2} - 4 \cdot 2 (- 224) = 46

y_{1, 2} = \frac{- 18 \pm 46}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- 18 + 46}{2 \cdot 2}, y_{2} = \frac{- 18 - 46}{2 \cdot 2}

y = \frac{- 18 + 46}{2 \cdot 2} : 7

y = \frac{- 18 - 46}{2 \cdot 2} : - 16

The solutions to the quadratic equation are:

y = 7, y = - 16

w^{2} + 13 w = 20

y, x^{2} + y^{2} = 2 x y

5 x - 17 = (x + 3) (x - 3)

9 x = 4 - 2 x^{2}

3 m^{2} - 16 m - 2 = - 7

What are the solutions to the equation (9+y)^2+y^2=305 ?
The solutions to the equation (9+y)^2+y^2=305 are y=7,y=-16
Find the zeros of (9+y)^2+y^2=305
The zeros of (9+y)^2+y^2=305 are y=7,y=-16