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

The solutions to the equation 2(x+1)^2-4=4 are x=1,x=-3

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

The zeros of 2(x+1)^2-4=4 are x=1,x=-3

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

x = 1, x = - 3

Solution steps

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

Move

4

to the right side

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

Divide both sides by

2

(x + 1)^{2} = 4

For

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

the solutions are

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

Solve

x + 1 = 4 : x = 1

Solve

x + 1 = - 4 : x = - 3

The solutions to the quadratic equation are:

x = 1, x = - 3

5 - 2 x \leq - 7

10 \cdot \frac{3}{5}

5 x = 35745

6 x^{2} + x

\frac{3 a}{2} - 8 = \frac{2 a - 7}{5}

What are the solutions to the equation 2(x+1)^2-4=4 ?
The solutions to the equation 2(x+1)^2-4=4 are x=1,x=-3
Find the zeros of 2(x+1)^2-4=4
The zeros of 2(x+1)^2-4=4 are x=1,x=-3