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2x(x+1)=3x(x-1)
We move all terms to the left:
2x(x+1)-(3x(x-1))=0
We multiply parentheses
2x^2+2x-(3x(x-1))=0
We calculate terms in parentheses: -(3x(x-1)), so:We get rid of parentheses
3x(x-1)
We multiply parentheses
3x^2-3x
Back to the equation:
-(3x^2-3x)
2x^2-3x^2+2x+3x=0
We add all the numbers together, and all the variables
-1x^2+5x=0
a = -1; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·(-1)·0
Δ = 25
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{25}=5$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*-1}=\frac{-10}{-2} =+5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*-1}=\frac{0}{-2} =0 $
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