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6x(x+2)=5(1+x)
We move all terms to the left:
6x(x+2)-(5(1+x))=0
We add all the numbers together, and all the variables
6x(x+2)-(5(x+1))=0
We multiply parentheses
6x^2+12x-(5(x+1))=0
We calculate terms in parentheses: -(5(x+1)), so:We get rid of parentheses
5(x+1)
We multiply parentheses
5x+5
Back to the equation:
-(5x+5)
6x^2+12x-5x-5=0
We add all the numbers together, and all the variables
6x^2+7x-5=0
a = 6; b = 7; c = -5;
Δ = b2-4ac
Δ = 72-4·6·(-5)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{169}=13$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-13}{2*6}=\frac{-20}{12} =-1+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+13}{2*6}=\frac{6}{12} =1/2 $
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