What are the solutions to the equation (2x+4)^2-4=12 ?

The solutions to the equation (2x+4)^2-4=12 are x=0,x=-4

Find the zeros of (2x+4)^2-4=12

The zeros of (2x+4)^2-4=12 are x=0,x=-4

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(2 x + 4)^{2} - 4 = 12

x = 0, x = - 4

Solution steps

(2 x + 4)^{2} - 4 = 12

Move

4

to the right side

(2 x + 4)^{2} = 16

For

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

the solutions are

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

Solve

2 x + 4 = 16 : x = 0

Solve

2 x + 4 = - 16 : x = - 4

The solutions to the quadratic equation are:

x = 0, x = - 4

u^{2} - 3 u = - \frac{5}{4}

x - 4 = - (- x + 4)

∣ x - 3 ∣ < - 15

(y - 4)^{2} = 7

x + (\frac{x}{3}) \leq 20

What are the solutions to the equation (2x+4)^2-4=12 ?
The solutions to the equation (2x+4)^2-4=12 are x=0,x=-4
Find the zeros of (2x+4)^2-4=12
The zeros of (2x+4)^2-4=12 are x=0,x=-4