What are the solutions to the equation -2(x+1)^2-4=12 ?

The solutions to the equation -2(x+1)^2-4=12 are x=2sqrt(2)i-1,x=-2sqrt(2)i-1

Find the zeros of -2(x+1)^2-4=12

The zeros of -2(x+1)^2-4=12 are x=2sqrt(2)i-1,x=-2sqrt(2)i-1

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- 2 (x + 1)^{2} - 4 = 12

x = 22 i - 1, x = - 22 i - 1

Solution steps

- 2 (x + 1)^{2} - 4 = 12

Move

4

to the right side

- 2 (x + 1)^{2} = 16

Divide both sides by

- 2

(x + 1)^{2} = - 8

For

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

the solutions are

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

Solve

x + 1 = - 8 : x = 22 i - 1

Solve

x + 1 = - - 8 : x = - 22 i - 1

The solutions to the quadratic equation are:

x = 22 i - 1, x = - 22 i - 1

- 6 x^{2} + 4 = 0

4^{2} = x^{2} + 3^{2}

0 = \frac{2}{3} x^{2} - 6 x + 12

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

(x + 4)^{2} = 13

What are the solutions to the equation -2(x+1)^2-4=12 ?
The solutions to the equation -2(x+1)^2-4=12 are x=2sqrt(2)i-1,x=-2sqrt(2)i-1
Find the zeros of -2(x+1)^2-4=12
The zeros of -2(x+1)^2-4=12 are x=2sqrt(2)i-1,x=-2sqrt(2)i-1