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