What are the solutions to the equation (y+14)^2+y^2-25=0 ?

The solutions to the equation (y+14)^2+y^2-25=0 are y=-7+isqrt(73/2),y=-7-isqrt(73/2)

Find the zeros of (y+14)^2+y^2-25=0

The zeros of (y+14)^2+y^2-25=0 are y=-7+isqrt(73/2),y=-7-isqrt(73/2)

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(y + 14)^{2} + y^{2} - 25 = 0

y = - 7 + i \frac{73}{2}, y = - 7 - i \frac{73}{2}

Solution steps

(y + 14)^{2} + y^{2} - 25 = 0

Expand

(y + 14)^{2} + y^{2} - 25 : 2 y^{2} + 28 y + 171

2 y^{2} + 28 y + 171 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 28 \pm 2 8 ^{2} - 4 \cdot 2 \cdot 171}{2 \cdot 2}

Simplify

2 8^{2} - 4 \cdot 2 \cdot 171 : 2146 i

y_{1, 2} = \frac{- 28 \pm 2 146 i}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- 28 + 2 146 i}{2 \cdot 2}, y_{2} = \frac{- 28 - 2 146 i}{2 \cdot 2}

y = \frac{- 28 + 2 146 i}{2 \cdot 2} : - 7 + i \frac{73}{2}

y = \frac{- 28 - 2 146 i}{2 \cdot 2} : - 7 - i \frac{73}{2}

The solutions to the quadratic equation are:

y = - 7 + i \frac{73}{2}, y = - 7 - i \frac{73}{2}

y, x^{2} + y^{2} = - 5

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

a (x) = (x - 3) (4 x + 1) + 2 (3 - x) (x + 2)

- 12 x^{2} = 720

m^{2} + 12 m = 0

What are the solutions to the equation (y+14)^2+y^2-25=0 ?
The solutions to the equation (y+14)^2+y^2-25=0 are y=-7+isqrt(73/2),y=-7-isqrt(73/2)
Find the zeros of (y+14)^2+y^2-25=0
The zeros of (y+14)^2+y^2-25=0 are y=-7+isqrt(73/2),y=-7-isqrt(73/2)