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

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

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

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

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(x + 2)^{2} = 4

x = 0, x = - 4

Solution steps

(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

2 a^{2} + 11 a + 5

\frac{2 ^{0}}{( 2 ^{- 2} x ^{\frac{1}{2}} y ^{- 2} ) ^{3}}

\frac{\frac{2}{3}}{\frac{1}{5}}

- \frac{2}{3} - 1

x > 3 and x > 6

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