What are the solutions to the equation (3x-7)(2x+5)=0 ?

The solutions to the equation (3x-7)(2x+5)=0 are x= 7/3 ,x=-5/2

Find the zeros of (3x-7)(2x+5)=0

The zeros of (3x-7)(2x+5)=0 are x= 7/3 ,x=-5/2

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(3 x - 7) (2 x + 5) = 0

x = \frac{7}{3}, x = - \frac{5}{2}

Solution steps

(3 x - 7) (2 x + 5) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

3 x - 7 = 0 or 2 x + 5 = 0

Solve

3 x - 7 = 0 : x = \frac{7}{3}

Solve

2 x + 5 = 0 : x = - \frac{5}{2}

The solutions to the quadratic equation are:

x = \frac{7}{3}, x = - \frac{5}{2}

lo g_{2} (16) = b

18 x^{2} - 21 x - 9

\frac{2}{5} m - \frac{4}{5} - \frac{3}{5} m

\frac{8 c + 25}{6 c} + \frac{5 - 2 c}{6 c}

(\frac{5}{8})^{4}

What are the solutions to the equation (3x-7)(2x+5)=0 ?
The solutions to the equation (3x-7)(2x+5)=0 are x= 7/3 ,x=-5/2
Find the zeros of (3x-7)(2x+5)=0
The zeros of (3x-7)(2x+5)=0 are x= 7/3 ,x=-5/2