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

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

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

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

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

x = 0, x = - 4

Solution steps

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

Move

4

to the right side

(x + 2)^{2} = 4

For

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

the solutions are

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

Solve

x + 2 = 4 : x = 0

Solve

x + 2 = - 4 : x = - 4

The solutions to the quadratic equation are:

x = 0, x = - 4

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

21 x^{2} - 32 x - 5

3 < 2 x - 3 < 15

∣ q + 7 ∣ \leq 9

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