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3(2x+4)=2x(3x+6)
We move all terms to the left:
3(2x+4)-(2x(3x+6))=0
We multiply parentheses
6x-(2x(3x+6))+12=0
We calculate terms in parentheses: -(2x(3x+6)), so:We get rid of parentheses
2x(3x+6)
We multiply parentheses
6x^2+12x
Back to the equation:
-(6x^2+12x)
-6x^2+6x-12x+12=0
We add all the numbers together, and all the variables
-6x^2-6x+12=0
a = -6; b = -6; c = +12;
Δ = b2-4ac
Δ = -62-4·(-6)·12
Δ = 324
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{324}=18$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-18}{2*-6}=\frac{-12}{-12} =1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+18}{2*-6}=\frac{24}{-12} =-2 $
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