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

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

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

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

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

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

Decimal Notation

x = - 0.5, x = 1.33333 \dots

Solution steps

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

2 x + 1 = 0 : x = - \frac{1}{2}

Solve

4 - 3 x = 0 : x = \frac{4}{3}

The solutions to the quadratic equation are:

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

x - 3 < 7

2 (\frac{15}{17}) (\frac{8}{17})

\frac{4 m}{4 m ^{2} - 1} - \frac{2 m + 1}{6 m - 3} + \frac{2 m - 1}{4 m + 2}

(\frac{12}{7})^{2}

\frac{5}{18} - \frac{16}{63} + \frac{1}{12}

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