What are the solutions to the equation -(x+1)^2-4=0 ?

The solutions to the equation -(x+1)^2-4=0 are x=2i-1,x=-2i-1

Find the zeros of -(x+1)^2-4=0

The zeros of -(x+1)^2-4=0 are x=2i-1,x=-2i-1

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- (x + 1)^{2} - 4 = 0

x = 2 i - 1, x = - 2 i - 1

Solution steps

- (x + 1)^{2} - 4 = 0

Move

4

to the right side

- (x + 1)^{2} = 4

Divide both sides by

- 1

(x + 1)^{2} = - 4

For

(g (x))^{2} = f (a)

the solutions are

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

Solve

x + 1 = - 4 : x = 2 i - 1

Solve

x + 1 = - - 4 : x = - 2 i - 1

The solutions to the quadratic equation are:

x = 2 i - 1, x = - 2 i - 1

\frac{1}{lo g _{b} ( 4 )}

3 (x^{2} - 3) + 1 = - 5

3 x - 2 = x + 1

\frac{81}{36}

x^{4} - 5 x^{2} + 4

What are the solutions to the equation -(x+1)^2-4=0 ?
The solutions to the equation -(x+1)^2-4=0 are x=2i-1,x=-2i-1
Find the zeros of -(x+1)^2-4=0
The zeros of -(x+1)^2-4=0 are x=2i-1,x=-2i-1