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

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

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

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

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

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

Solution steps

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

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

Solve

4 x - 1 = 0 : x = \frac{1}{4}

The solutions to the quadratic equation are:

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

n^{2} + 3 n - 10

2 v^{3} - 2 v^{3} x^{4}

\frac{5}{2 t - 1} < 2

x^{3} - 3 x - 52

5 x y^{2} (2 y - 3 x) - 15 x^{2} (3 x - 2 y)^{2}

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