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12x(2x-2)=3x(x+4)
We move all terms to the left:
12x(2x-2)-(3x(x+4))=0
We multiply parentheses
24x^2-24x-(3x(x+4))=0
We calculate terms in parentheses: -(3x(x+4)), so:We get rid of parentheses
3x(x+4)
We multiply parentheses
3x^2+12x
Back to the equation:
-(3x^2+12x)
24x^2-3x^2-24x-12x=0
We add all the numbers together, and all the variables
21x^2-36x=0
a = 21; b = -36; c = 0;
Δ = b2-4ac
Δ = -362-4·21·0
Δ = 1296
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{1296}=36$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-36}{2*21}=\frac{0}{42} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+36}{2*21}=\frac{72}{42} =1+5/7 $
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