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Softa [21]
3 years ago
13

Solve (y-4)^2+63=0 where y is a real number.

Mathematics
1 answer:
sergeinik [125]3 years ago
4 0

Answer:

y = 4 + 3 \sqrt{7} i

Step-by-step explanation:

(y-4)^2+63=0  \\  \\ (y-4)^2 =  - 63 \\  \\ (y-4)=  \sqrt{ - 63}  \\  \\ y - 4 = 3 \sqrt{7} i \\  \\ y = 4 + 3 \sqrt{7} i

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