What are the solutions to the equation (3n-2)(4n+1)=0 ?

The solutions to the equation (3n-2)(4n+1)=0 are n= 2/3 ,n=-1/4

Find the zeros of (3n-2)(4n+1)=0

The zeros of (3n-2)(4n+1)=0 are n= 2/3 ,n=-1/4

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(3 n - 2) (4 n + 1) = 0

n = \frac{2}{3}, n = - \frac{1}{4}

Decimal Notation

n = 0.66666 \dots, n = - 0.25

Solution steps

(3 n - 2) (4 n + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

3 n - 2 = 0 or 4 n + 1 = 0

Solve

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

Solve

4 n + 1 = 0 : n = - \frac{1}{4}

The solutions to the quadratic equation are:

n = \frac{2}{3}, n = - \frac{1}{4}

2 x \cdot 4 x^{\frac{- 5}{2}}

128 - 32 x + 2 x^{2}

- 5 + 14

x (x - 3) = 28

6 x - 1 = 0

What are the solutions to the equation (3n-2)(4n+1)=0 ?
The solutions to the equation (3n-2)(4n+1)=0 are n= 2/3 ,n=-1/4
Find the zeros of (3n-2)(4n+1)=0
The zeros of (3n-2)(4n+1)=0 are n= 2/3 ,n=-1/4