What are the solutions to the equation (3x+1)(2x-5)=0 ?

The solutions to the equation (3x+1)(2x-5)=0 are x=-1/3 ,x= 5/2

Find the zeros of (3x+1)(2x-5)=0

The zeros of (3x+1)(2x-5)=0 are x=-1/3 ,x= 5/2

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(3 x + 1) (2 x - 5) = 0

x = - \frac{1}{3}, x = \frac{5}{2}

Decimal Notation

x = - 0.33333 \dots, x = 2.5

Solution steps

(3 x + 1) (2 x - 5) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

3 x + 1 = 0 or 2 x - 5 = 0

Solve

3 x + 1 = 0 : x = - \frac{1}{3}

Solve

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

The solutions to the quadratic equation are:

x = - \frac{1}{3}, x = \frac{5}{2}

25 x^{2} - 9 x = 0

x^{2} = - 361

0 = - 0.06 x^{2} + 1.31 x + 1.38

5 x^{2} - 20 x - 5

600 x + 400 x (x + 200) = 1700000

What are the solutions to the equation (3x+1)(2x-5)=0 ?
The solutions to the equation (3x+1)(2x-5)=0 are x=-1/3 ,x= 5/2
Find the zeros of (3x+1)(2x-5)=0
The zeros of (3x+1)(2x-5)=0 are x=-1/3 ,x= 5/2