What are the solutions to the equation -4(x+2)^2+16=0 ?

The solutions to the equation -4(x+2)^2+16=0 are x=0,x=-4

Find the zeros of -4(x+2)^2+16=0

The zeros of -4(x+2)^2+16=0 are x=0,x=-4

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- 4 (x + 2)^{2} + 16 = 0

x = 0, x = - 4

Solution steps

- 4 (x + 2)^{2} + 16 = 0

Move

16

to the right side

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

Divide both sides by

- 4

(x + 2)^{2} = 4

For

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

the solutions are

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

Solve

x + 2 = 4 : x = 0

Solve

x + 2 = - 4 : x = - 4

The solutions to the quadratic equation are:

x = 0, x = - 4

4 q^{2} + q + 1 = - 3 q

x^{2} + x - 20 = 10

x^{2} + 35 = 14 x

72 + 6 x = x^{2}

y^{2} + 8 y + 20 = 0

What are the solutions to the equation -4(x+2)^2+16=0 ?
The solutions to the equation -4(x+2)^2+16=0 are x=0,x=-4
Find the zeros of -4(x+2)^2+16=0
The zeros of -4(x+2)^2+16=0 are x=0,x=-4