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

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

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

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

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(3 j + 5) (2 j - 3) = 0

j = - \frac{5}{3}, j = \frac{3}{2}

Decimal Notation

j = - 1.66666 \dots, j = 1.5

Solution steps

(3 j + 5) (2 j - 3) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

3 j + 5 = 0 or 2 j - 3 = 0

Solve

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

Solve

2 j - 3 = 0 : j = \frac{3}{2}

The solutions to the quadratic equation are:

j = - \frac{5}{3}, j = \frac{3}{2}

10 q^{2} = q + 2

z^{2} + z - 3 = 0

3 x^{2} - 45 = 2 x^{2}

3 x = x^{2} - 2

9 x^{2} + 6 x = - 10

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