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

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

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

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

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(x + 1)^{2} = 2

x = 2 - 1, x = - 2 - 1

Decimal Notation

x = 0.41421 \dots, x = - 2.41421 \dots

Solution steps

(x + 1)^{2} = 2

For

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

the solutions are

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

Solve

x + 1 = 2 : x = 2 - 1

Solve

x + 1 = - 2 : x = - 2 - 1

The solutions to the quadratic equation are:

x = 2 - 1, x = - 2 - 1

x^{2} + 2 x

∣ x + 6 ∣ \leq 2

3 x^{2} + 6 x - 15

6 c^{2} + 10 c^{3}

9 x^{2} - 4 x - 1 = 0

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