What are the solutions to the equation x^2-4=sqrt(-7) ?

The solutions to the equation x^2-4=sqrt(-7) are

Find the zeros of x^2-4=sqrt(-7)

The zeros of x^2-4=sqrt(-7) are

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x^{2} - 4 = - 7

x = 423 cos \frac{arctan ( \frac{7}{4} )}{2} + 423 i sin \frac{arctan ( \frac{7}{4} )}{2}, x = 423 cos \frac{arctan ( \frac{7}{4} ) + 2 π}{2} + 423 i sin \frac{arctan ( \frac{7}{4} ) + 2 π}{2}

Solution steps

x^{2} - 4 = - 7

Expand

- 7 : 7 i

x^{2} - 4 = 7 i

Move

4

to the right side

x^{2} = 7 i + 4

For

z^{n} = a

the solutions are

z_{k} = n ∣ a ∣ (cos (\frac{ar g ( a ) + 2 kπ}{n}) + i sin (\frac{ar g ( a ) + 2 kπ}{n})),

k = 0, 1, \dots, n - 1

For

n = 2, a = 7 i + 4

∣ a ∣ = 23

ar g (a) = arctan (\frac{7}{4})

x = 23 cos \frac{arctan ( \frac{7}{4} ) + 2 \cdot 0 π}{2} + i sin \frac{arctan ( \frac{7}{4} ) + 2 \cdot 0 π}{2}, x = 23 cos \frac{arctan ( \frac{7}{4} ) + 2 \cdot 1 π}{2} + i sin \frac{arctan ( \frac{7}{4} ) + 2 \cdot 1 π}{2}

Simplify

23 cos \frac{arctan ( \frac{7}{4} ) + 2 \cdot 0 π}{2} + i sin \frac{arctan ( \frac{7}{4} ) + 2 \cdot 0 π}{2} : 423 cos \frac{arctan ( \frac{7}{4} )}{2} + 423 i sin \frac{arctan ( \frac{7}{4} )}{2}

Simplify

23 cos \frac{arctan ( \frac{7}{4} ) + 2 \cdot 1 π}{2} + i sin \frac{arctan ( \frac{7}{4} ) + 2 \cdot 1 π}{2} : 423 cos \frac{arctan ( \frac{7}{4} ) + 2 π}{2} + 423 i sin \frac{arctan ( \frac{7}{4} ) + 2 π}{2}

x = 423 cos \frac{arctan ( \frac{7}{4} )}{2} + 423 i sin \frac{arctan ( \frac{7}{4} )}{2}, x = 423 cos \frac{arctan ( \frac{7}{4} ) + 2 π}{2} + 423 i sin \frac{arctan ( \frac{7}{4} ) + 2 π}{2}

25 x^{2} - 4 = 0

z^{2} + 4 z

x^{2} + (\frac{4 - x}{2})^{2} = 16

3 x^{2} = 81

m p^{2} + 2 m + n p^{2} = 2 n

What are the solutions to the equation x^2-4=sqrt(-7) ?
The solutions to the equation x^2-4=sqrt(-7) are
Find the zeros of x^2-4=sqrt(-7)
The zeros of x^2-4=sqrt(-7) are