3x2+17x+3=(x+1)(3x+13)

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Solution for 3x2+17x+3=(x+1)(3x+13) equation: 



3x^2+17x+3=(x+1)(3x+13)

We move all terms to the left:

3x^2+17x+3-((x+1)(3x+13))=0

We multiply parentheses ..

3x^2-((+3x^2+13x+3x+13))+17x+3=0



We calculate terms in parentheses: -((+3x^2+13x+3x+13)), so:

(+3x^2+13x+3x+13)

We get rid of parentheses

3x^2+13x+3x+13

We add all the numbers together, and all the variables

3x^2+16x+13

Back to the equation:

-(3x^2+16x+13)



We add all the numbers together, and all the variables

3x^2+17x-(3x^2+16x+13)+3=0

We get rid of parentheses

3x^2-3x^2+17x-16x-13+3=0

We add all the numbers together, and all the variables

x-10=0

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

x=10

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