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

The solutions to the equation (x+1)^2=12 are x=2sqrt(3)-1,x=-2sqrt(3)-1

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

The zeros of (x+1)^2=12 are x=2sqrt(3)-1,x=-2sqrt(3)-1

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(x + 1)^{2} = 12

x = 23 - 1, x = - 23 - 1

Decimal Notation

x = 2.46410 \dots, x = - 4.46410 \dots

Solution steps

(x + 1)^{2} = 12

For

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

the solutions are

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

Solve

x + 1 = 12 : x = 23 - 1

Solve

x + 1 = - 12 : x = - 23 - 1

The solutions to the quadratic equation are:

x = 23 - 1, x = - 23 - 1

x (1 - x^{2})^{3} > 3 (1 - x^{2})^{3}

0 = x^{2} + x - 30

\frac{18}{25}

x^{2} + 5 x + 4 \geq 0

2 x - 3 < 1 + x < 3 x - 1

What are the solutions to the equation (x+1)^2=12 ?
The solutions to the equation (x+1)^2=12 are x=2sqrt(3)-1,x=-2sqrt(3)-1
Find the zeros of (x+1)^2=12
The zeros of (x+1)^2=12 are x=2sqrt(3)-1,x=-2sqrt(3)-1