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

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

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

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

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

x = 2 - 4, x = - 2 - 4

Solution steps

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

Switch sides

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

Move

2

to the right side

(x + 4)^{2} = 2

For

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

the solutions are

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

Solve

x + 4 = 2 : x = 2 - 4

Solve

x + 4 = - 2 : x = - 2 - 4

The solutions to the quadratic equation are:

x = 2 - 4, x = - 2 - 4

19 \leq 5 x - 1 \leq 39

∣ x + 3 ∣ > 8

7 x^{2} - 20 x + 4 = 8 x - 10

m^{2} - 3 m - 10

2 x - 5 = x + 1

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