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

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

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

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

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

x = - 1, x = - 3

Solution steps

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

Move

4

to the right side

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

Divide both sides by

4

(x + 2)^{2} = 1

For

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

the solutions are

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

Solve

x + 2 = 1 : x = - 1

Solve

x + 2 = - 1 : x = - 3

The solutions to the quadratic equation are:

x = - 1, x = - 3

x, 3 y^{2} + 5 x y = 2 x^{2}

3 q^{2} - 108 q + 890 = 0

x (2 x - 7) = 15

x^{2} - 9 x = 7

7 x^{2} - 78 x = 0

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