What are the solutions to the equation 1+(2+x)^2-4=0 ?

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

Find the zeros of 1+(2+x)^2-4=0

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

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1 + (2 + x)^{2} - 4 = 0

x = 3 - 2, x = - 3 - 2

Decimal Notation

x = - 0.26794 \dots, x = - 3.73205 \dots

Solution steps

1 + (2 + x)^{2} - 4 = 0

Move

1

to the right side

(2 + x)^{2} - 4 = - 1

Move

4

to the right side

(2 + x)^{2} = 3

For

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

the solutions are

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

Solve

2 + x = 3 : x = 3 - 2

Solve

2 + x = - 3 : x = - 3 - 2

The solutions to the quadratic equation are:

x = 3 - 2, x = - 3 - 2

6 x^{2} - 1944 = 0

x^{2} + 2 x = 120

c, A = \frac{1}{6} b c^{2}

5 x^{2} + 10 x = - 5

z^{2} + 11 z + 24 = 0

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