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(6-x)(2x)=(x+3)(x)
We move all terms to the left:
(6-x)(2x)-((x+3)(x))=0
We add all the numbers together, and all the variables
(-1x+6)2x-((x+3)x)=0
We multiply parentheses
-2x^2+12x-((x+3)x)=0
We calculate terms in parentheses: -((x+3)x), so:We get rid of parentheses
(x+3)x
We multiply parentheses
x^2+3x
Back to the equation:
-(x^2+3x)
-2x^2-x^2+12x-3x=0
We add all the numbers together, and all the variables
-3x^2+9x=0
a = -3; b = 9; c = 0;
Δ = b2-4ac
Δ = 92-4·(-3)·0
Δ = 81
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{81}=9$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-9}{2*-3}=\frac{-18}{-6} =+3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+9}{2*-3}=\frac{0}{-6} =0 $
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