What are the solutions to the equation -x^2+3=124 ?

The solutions to the equation -x^2+3=124 are x=11i,x=-11i

Find the zeros of -x^2+3=124

The zeros of -x^2+3=124 are x=11i,x=-11i

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- x^{2} + 3 = 124

x = 11 i, x = - 11 i

Solution steps

- x^{2} + 3 = 124

Move

3

to the right side

- x^{2} = 121

Divide both sides by

- 1

x^{2} = - 121

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

x = - 121, x = - - 121

- 121 = 11 i

- - 121 = - 11 i

x = 11 i, x = - 11 i

2 = 3 x + 6

73 0^{\frac{3}{2}}

3 b^{2} - 4 b - 4 = 0

x^{2} - 12 x + 11 \leq 0

6 g^{3} + 12 g^{2} + 17 g + 34

What are the solutions to the equation -x^2+3=124 ?
The solutions to the equation -x^2+3=124 are x=11i,x=-11i
Find the zeros of -x^2+3=124
The zeros of -x^2+3=124 are x=11i,x=-11i