What are the solutions to the equation (x+12)^2=1 ?

The solutions to the equation (x+12)^2=1 are x=-11,x=-13

Find the zeros of (x+12)^2=1

The zeros of (x+12)^2=1 are x=-11,x=-13

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(x + 12)^{2} = 1

x = - 11, x = - 13

Solution steps

(x + 12)^{2} = 1

For

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

the solutions are

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

Solve

x + 12 = 1 : x = - 11

Solve

x + 12 = - 1 : x = - 13

The solutions to the quadratic equation are:

x = - 11, x = - 13

∣ x - 1 ∣ = 5

\frac{x}{x - 3} < 4

x^{2} - 8 x - 3 = 0

(12 x - y)^{2} - (4 x - 3 y)^{2}

7 x - 21 = x + 15

What are the solutions to the equation (x+12)^2=1 ?
The solutions to the equation (x+12)^2=1 are x=-11,x=-13
Find the zeros of (x+12)^2=1
The zeros of (x+12)^2=1 are x=-11,x=-13