j(2j+3)+20=2j(j-3)

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Solution for j(2j+3)+20=2j(j-3) equation: 



j(2j+3)+20=2j(j-3)

We move all terms to the left:

j(2j+3)+20-(2j(j-3))=0

We multiply parentheses

2j^2+3j-(2j(j-3))+20=0



We calculate terms in parentheses: -(2j(j-3)), so:

2j(j-3)

We multiply parentheses

2j^2-6j

Back to the equation:

-(2j^2-6j)



We get rid of parentheses

2j^2-2j^2+3j+6j+20=0

We add all the numbers together, and all the variables

9j+20=0

We move all terms containing j to the left, all other terms to the right

9j=-20

j=-20/9

j=-2+2/9

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