What are the solutions to the equation 0^2+y^2-14(0)+4y+28=0 ?

The solutions to the equation 0^2+y^2-14(0)+4y+28=0 are y=-2+2sqrt(6)i,y=-2-2sqrt(6)i

Find the zeros of 0^2+y^2-14(0)+4y+28=0

The zeros of 0^2+y^2-14(0)+4y+28=0 are y=-2+2sqrt(6)i,y=-2-2sqrt(6)i

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0^{2} + y^{2} - 14 (0) + 4 y + 28 = 0

y = - 2 + 26 i, y = - 2 - 26 i

Solution steps

0^{2} + y^{2} - 14 (0) + 4 y + 28 = 0

Expand

0^{2} + y^{2} - 14 (0) + 4 y + 28 : y^{2} + 4 y + 28

y^{2} + 4 y + 28 = 0

Solve with the quadratic formula

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

Simplify

4^{2} - 4 \cdot 1 \cdot 28 : 46 i

y_{1, 2} = \frac{- 4 \pm 4 6 i}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 4 + 4 6 i}{2 \cdot 1}, y_{2} = \frac{- 4 - 4 6 i}{2 \cdot 1}

y = \frac{- 4 + 4 6 i}{2 \cdot 1} : - 2 + 26 i

y = \frac{- 4 - 4 6 i}{2 \cdot 1} : - 2 - 26 i

The solutions to the quadratic equation are:

y = - 2 + 26 i, y = - 2 - 26 i

(x - y)^{2} + 3 (y - x)^{3}

15 a c - 9 a d - 30 b c + 18 b d

8 x^{3} + 125 y^{3}

∣ 25 x - 8 ∣ > 7

- 3 + 3

What are the solutions to the equation 0^2+y^2-14(0)+4y+28=0 ?
The solutions to the equation 0^2+y^2-14(0)+4y+28=0 are y=-2+2sqrt(6)i,y=-2-2sqrt(6)i
Find the zeros of 0^2+y^2-14(0)+4y+28=0
The zeros of 0^2+y^2-14(0)+4y+28=0 are y=-2+2sqrt(6)i,y=-2-2sqrt(6)i