What are the solutions to the equation (5+y)^2+y^2=37 ?

The solutions to the equation (5+y)^2+y^2=37 are y=1,y=-6

Find the zeros of (5+y)^2+y^2=37

The zeros of (5+y)^2+y^2=37 are y=1,y=-6

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(5 + y)^{2} + y^{2} = 37

y = 1, y = - 6

Solution steps

(5 + y)^{2} + y^{2} = 37

Expand

(5 + y)^{2} + y^{2} : 25 + 10 y + 2 y^{2}

25 + 10 y + 2 y^{2} = 37

Move

37

to the left side

2 y^{2} + 10 y - 12 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 2 ( - 12 )}{2 \cdot 2}

1 0^{2} - 4 \cdot 2 (- 12) = 14

y_{1, 2} = \frac{- 10 \pm 14}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- 10 + 14}{2 \cdot 2}, y_{2} = \frac{- 10 - 14}{2 \cdot 2}

y = \frac{- 10 + 14}{2 \cdot 2} : 1

y = \frac{- 10 - 14}{2 \cdot 2} : - 6

The solutions to the quadratic equation are:

y = 1, y = - 6

x^{2} + 6 x = - 5

x, f = 2 x^{2} - 2 y

- 16 t^{2} + 1600 = 0

8 (3 x^{2} + 4) = 0

x, x^{2} - x y + y^{2} = 27

What are the solutions to the equation (5+y)^2+y^2=37 ?
The solutions to the equation (5+y)^2+y^2=37 are y=1,y=-6
Find the zeros of (5+y)^2+y^2=37
The zeros of (5+y)^2+y^2=37 are y=1,y=-6