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

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

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

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

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(x + 1)^{2} = 4

x = 1, x = - 3

Solution steps

(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

- \frac{5}{9} y + 6 = 4 - \frac{4}{3} y

5 x^{2} + 2 x + 3 \leq 0

3 x + 6 = 21

3 x^{2} + 2 x - 16

5 - 3 x > - 19

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