3x=(2x+5)(4+x)

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Solution for 3x=(2x+5)(4+x) equation: 



3x=(2x+5)(4+x)

We move all terms to the left:

3x-((2x+5)(4+x))=0

We add all the numbers together, and all the variables

3x-((2x+5)(x+4))=0

We multiply parentheses ..

-((+2x^2+8x+5x+20))+3x=0



We calculate terms in parentheses: -((+2x^2+8x+5x+20)), so:

(+2x^2+8x+5x+20)

We get rid of parentheses

2x^2+8x+5x+20

We add all the numbers together, and all the variables

2x^2+13x+20

Back to the equation:

-(2x^2+13x+20)



We add all the numbers together, and all the variables

3x-(2x^2+13x+20)=0

We get rid of parentheses

-2x^2+3x-13x-20=0

We add all the numbers together, and all the variables

-2x^2-10x-20=0

a = -2; b = -10; c = -20;
Δ = b2-4ac
Δ = -102-4·(-2)·(-20)
Δ = -60
Delta is less than zero, so there is no solution for the equation

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