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

The solutions to the equation (x+2)^2-1=4 are x=sqrt(5)-2,x=-sqrt(5)-2

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

The zeros of (x+2)^2-1=4 are x=sqrt(5)-2,x=-sqrt(5)-2

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

x = 5 - 2, x = - 5 - 2

Decimal Notation

x = 0.23606 \dots, x = - 4.23606 \dots

Solution steps

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

Move

1

to the right side

(x + 2)^{2} = 5

For

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

the solutions are

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

Solve

x + 2 = 5 : x = 5 - 2

Solve

x + 2 = - 5 : x = - 5 - 2

The solutions to the quadratic equation are:

x = 5 - 2, x = - 5 - 2

0 = - 24 x^{2} + 16

x, f = x^{2} + k x + 2

x^{2} + 2 x - 100 = 0

x^{2} + 12 x + 3 = 7

5 b^{2} + b - 2 = 0

What are the solutions to the equation (x+2)^2-1=4 ?
The solutions to the equation (x+2)^2-1=4 are x=sqrt(5)-2,x=-sqrt(5)-2
Find the zeros of (x+2)^2-1=4
The zeros of (x+2)^2-1=4 are x=sqrt(5)-2,x=-sqrt(5)-2