What are the solutions to the equation 3(2x+3)(5x-4)=0 ?

The solutions to the equation 3(2x+3)(5x-4)=0 are x=-3/2 ,x= 4/5

Find the zeros of 3(2x+3)(5x-4)=0

The zeros of 3(2x+3)(5x-4)=0 are x=-3/2 ,x= 4/5

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3 (2 x + 3) (5 x - 4) = 0

x = - \frac{3}{2}, x = \frac{4}{5}

Solution steps

3 (2 x + 3) (5 x - 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

2 x + 3 = 0 or 5 x - 4 = 0

Solve

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

Solve

5 x - 4 = 0 : x = \frac{4}{5}

The solutions to the quadratic equation are:

x = - \frac{3}{2}, x = \frac{4}{5}

27 n = 216

2 - \frac{1}{2}

\frac{75}{0.15} = \frac{0.15 \cdot x}{0.15}

\frac{x - 6}{x + 4} \geq 0

\frac{3}{x + 2} > \frac{1}{x - 1}

What are the solutions to the equation 3(2x+3)(5x-4)=0 ?
The solutions to the equation 3(2x+3)(5x-4)=0 are x=-3/2 ,x= 4/5
Find the zeros of 3(2x+3)(5x-4)=0
The zeros of 3(2x+3)(5x-4)=0 are x=-3/2 ,x= 4/5