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

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

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

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

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(n + 2) (n - 2) = 0

n = - 2, n = 2

Solution steps

(n + 2) (n - 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

n + 2 = 0 or n - 2 = 0

Solve

n + 2 = 0 : n = - 2

Solve

n - 2 = 0 : n = 2

The solutions to the quadratic equation are:

n = - 2, n = 2

5 x^{2} - 3 x

12 c^{2} + 60 c + 75

\frac{4 + 10 i}{3 + i}

a^{2} - 14 ab + 48 b^{2}

4 n^{2} - 12 n + 9 = 0

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