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

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

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

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

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

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

Solution steps

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

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

Solve

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

The solutions to the quadratic equation are:

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

2 + 2 f \geq 10

3 (\frac{8}{9})^{0}

\frac{( m ^{6} - n ^{6} )}{( m - n )}

\frac{6 n ^{2} + 8 n + 2}{2 n ^{2} - 2}

5 \cdot 3 \cdot 2

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