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

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

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

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

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

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

Decimal Notation

g = - 0.33333 \dots, g = 2.5

Solution steps

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

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

Solve

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

The solutions to the quadratic equation are:

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

6 w^{2} - 30 w = 0

x^{2} + 3 x + 5 = 15

x^{2} - 5 = 59

x^{2} - 5 = 13

x^{2} - 3.5 x + 3 = 0

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