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

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

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

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

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

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

Decimal Notation

x = 0.5, x = - 1.66666 \dots

Solution steps

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

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

Solve

3 x + 5 = 0 : x = - \frac{5}{3}

The solutions to the quadratic equation are:

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

(5^{\frac{1}{3}})^{\frac{1}{3}}

16 x^{5} - 81 x

z^{2} - 10 z = - 17

6 \cdot 8

48 x - 16 x^{2} = 0

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