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3x-9=3x(x-3)
We move all terms to the left:
3x-9-(3x(x-3))=0
We calculate terms in parentheses: -(3x(x-3)), so:We get rid of parentheses
3x(x-3)
We multiply parentheses
3x^2-9x
Back to the equation:
-(3x^2-9x)
-3x^2+3x+9x-9=0
We add all the numbers together, and all the variables
-3x^2+12x-9=0
a = -3; b = 12; c = -9;
Δ = b2-4ac
Δ = 122-4·(-3)·(-9)
Δ = 36
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{36}=6$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6}{2*-3}=\frac{-18}{-6} =+3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6}{2*-3}=\frac{-6}{-6} =1 $
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