What are the solutions to the equation (x+1)^2-12=0 ?

The solutions to the equation (x+1)^2-12=0 are x=2sqrt(3)-1,x=-2sqrt(3)-1

Find the zeros of (x+1)^2-12=0

The zeros of (x+1)^2-12=0 are x=2sqrt(3)-1,x=-2sqrt(3)-1

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(x + 1)^{2} - 12 = 0

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

Decimal Notation

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

Solution steps

(x + 1)^{2} - 12 = 0

Move

12

to the right side

(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^{2} - 9995 x - 50000 = 0

m^{2} + 16 m = 4

x^{2} + 19 x + 48 = 0

3 x^{3} + 8 = 9 x^{2} + 3 x^{3}

(x + 11)^{2} - 225 = 0

What are the solutions to the equation (x+1)^2-12=0 ?
The solutions to the equation (x+1)^2-12=0 are x=2sqrt(3)-1,x=-2sqrt(3)-1
Find the zeros of (x+1)^2-12=0
The zeros of (x+1)^2-12=0 are x=2sqrt(3)-1,x=-2sqrt(3)-1