What are the solutions to the equation 1+2(x+3)^2=17 ?

The solutions to the equation 1+2(x+3)^2=17 are x=2sqrt(2)-3,x=-2sqrt(2)-3

Find the zeros of 1+2(x+3)^2=17

The zeros of 1+2(x+3)^2=17 are x=2sqrt(2)-3,x=-2sqrt(2)-3

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1 + 2 (x + 3)^{2} = 17

x = 22 - 3, x = - 22 - 3

Decimal Notation

x = - 0.17157 \dots, x = - 5.82842 \dots

Solution steps

1 + 2 (x + 3)^{2} = 17

Move

1

to the right side

2 (x + 3)^{2} = 16

Divide both sides by

2

(x + 3)^{2} = 8

For

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

the solutions are

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

Solve

x + 3 = 8 : x = 22 - 3

Solve

x + 3 = - 8 : x = - 22 - 3

The solutions to the quadratic equation are:

x = 22 - 3, x = - 22 - 3

x^{2} = - 5 x + 6

26 = 6 + 37 t - 16 t^{2}

7 x^{2} + 2 x + 1 = 0

x^{2} = - 4 x + 8

3 x^{2} - 2 = 190

What are the solutions to the equation 1+2(x+3)^2=17 ?
The solutions to the equation 1+2(x+3)^2=17 are x=2sqrt(2)-3,x=-2sqrt(2)-3
Find the zeros of 1+2(x+3)^2=17
The zeros of 1+2(x+3)^2=17 are x=2sqrt(2)-3,x=-2sqrt(2)-3