What are the solutions to the equation 2(x+1)^2-17=27 ?

The solutions to the equation 2(x+1)^2-17=27 are x=sqrt(22)-1,x=-sqrt(22)-1

Find the zeros of 2(x+1)^2-17=27

The zeros of 2(x+1)^2-17=27 are x=sqrt(22)-1,x=-sqrt(22)-1

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2 (x + 1)^{2} - 17 = 27

x = 22 - 1, x = - 22 - 1

Decimal Notation

x = 3.69041 \dots, x = - 5.69041 \dots

Solution steps

2 (x + 1)^{2} - 17 = 27

Move

17

to the right side

2 (x + 1)^{2} = 44

Divide both sides by

2

(x + 1)^{2} = 22

For

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

the solutions are

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

Solve

x + 1 = 22 : x = 22 - 1

Solve

x + 1 = - 22 : x = - 22 - 1

The solutions to the quadratic equation are:

x = 22 - 1, x = - 22 - 1

m^{2} + m - 2 = 0

5 (x + 9)^{2} + 41 = 46

y^{2} = 576

- 11 x^{2} = x + 11

(x - 8)^{2} = 144

What are the solutions to the equation 2(x+1)^2-17=27 ?
The solutions to the equation 2(x+1)^2-17=27 are x=sqrt(22)-1,x=-sqrt(22)-1
Find the zeros of 2(x+1)^2-17=27
The zeros of 2(x+1)^2-17=27 are x=sqrt(22)-1,x=-sqrt(22)-1