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

The solutions to the equation (2x+4)^2-4=0 are x=-1,x=-3

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

The zeros of (2x+4)^2-4=0 are x=-1,x=-3

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

x = - 1, x = - 3

Solution steps

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

Move

4

to the right side

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

For

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

the solutions are

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

Solve

2 x + 4 = 4 : x = - 1

Solve

2 x + 4 = - 4 : x = - 3

The solutions to the quadratic equation are:

x = - 1, x = - 3

x^{2} + 2 x + 1 = 0

4 (6 - 3 x)^{2} - 38 = - 37

m^{2} + 10 m + 25 = 0

x (2 x + 1) = 0

x, x^{2} + x - 2 = 0

What are the solutions to the equation (2x+4)^2-4=0 ?
The solutions to the equation (2x+4)^2-4=0 are x=-1,x=-3
Find the zeros of (2x+4)^2-4=0
The zeros of (2x+4)^2-4=0 are x=-1,x=-3