What are the solutions to the equation (3n+2)(n-2)=0 ?

The solutions to the equation (3n+2)(n-2)=0 are n=-2/3 ,n=2

Find the zeros of (3n+2)(n-2)=0

The zeros of (3n+2)(n-2)=0 are n=-2/3 ,n=2

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

n = - \frac{2}{3}, n = 2

Decimal Notation

n = - 0.66666 \dots, n = 2

Solution steps

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

3 n + 2 = 0 or n - 2 = 0

Solve

3 n + 2 = 0 : n = - \frac{2}{3}

Solve

n - 2 = 0 : n = 2

The solutions to the quadratic equation are:

n = - \frac{2}{3}, n = 2

- 2 x + 3 \leq 13

3 x - 1 + 5 x = 5 (x - 1)

(1 + \frac{i}{n})^{2}

5 x^{2} = - 3 x - 4

9^{5 n} \leq \frac{1}{6561}

What are the solutions to the equation (3n+2)(n-2)=0 ?
The solutions to the equation (3n+2)(n-2)=0 are n=-2/3 ,n=2
Find the zeros of (3n+2)(n-2)=0
The zeros of (3n+2)(n-2)=0 are n=-2/3 ,n=2