What are the solutions to the equation (x+2)^2-3=1 ?

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

Find the zeros of (x+2)^2-3=1

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

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(x + 2)^{2} - 3 = 1

x = 0, x = - 4

Solution steps

(x + 2)^{2} - 3 = 1

Move

3

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 - 5) = 6

9 m^{2} - 30 mn + 25 n^{2}

\frac{3}{8} \cdot 12

\frac{4 + - 81}{7 - - 64}

- \frac{1}{4} \cdot 2

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