What are the solutions to the equation -0.5(n+2)^2=0 ?

The solutions to the equation -0.5(n+2)^2=0 are n=-2

Find the zeros of -0.5(n+2)^2=0

The zeros of -0.5(n+2)^2=0 are n=-2

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- 0.5 (n + 2)^{2} = 0

n = - 2

Solution steps

- 0.5 (n + 2)^{2} = 0

Divide both sides by

- 0.5

(n + 2)^{2} = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

Solve

n + 2 = 0 : n = - 2

The solution to the quadratic equation is:

n = - 2

a^{3} + 2 a b^{2} - 4 b^{3} - 2 a^{2} b

(5 x y^{5})^{2} (y^{3})^{4}

1^{- 1}

36 \cdot 9^{\frac{3}{2}}

- 7 = - (- 4 + 3 n) + 2 (n - 2)

What are the solutions to the equation -0.5(n+2)^2=0 ?
The solutions to the equation -0.5(n+2)^2=0 are n=-2
Find the zeros of -0.5(n+2)^2=0
The zeros of -0.5(n+2)^2=0 are n=-2