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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:We add all the numbers together, and all the variables
(+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)
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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