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3x(x+1)=(x+1)
We move all terms to the left:
3x(x+1)-((x+1))=0
We multiply parentheses
3x^2+3x-((x+1))=0
We calculate terms in parentheses: -((x+1)), so:We get rid of parentheses
(x+1)
We get rid of parentheses
x+1
Back to the equation:
-(x+1)
3x^2+3x-x-1=0
We add all the numbers together, and all the variables
3x^2+2x-1=0
a = 3; b = 2; c = -1;
Δ = b2-4ac
Δ = 22-4·3·(-1)
Δ = 16
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{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4}{2*3}=\frac{-6}{6} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4}{2*3}=\frac{2}{6} =1/3 $
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