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

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

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

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

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

x = 0, x = - 4

Solution steps

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

Switch sides

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

Move

4

to the right side

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

Divide both sides by

- 1

(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

t, h = v t + \frac{1}{2} g t^{2}

x, 12 x^{2} - 8 x s + s^{2} = 0

- 3 x^{2} - 27 = 0

b^{2} = 121

y, (3 x y - x^{2})^{2} = 2 y + x y

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