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(x+9)(3x-3)=(2x)(3x+3)
We move all terms to the left:
(x+9)(3x-3)-((2x)(3x+3))=0
We multiply parentheses ..
(+3x^2-3x+27x-27)-(2x(3x+3))=0
We calculate terms in parentheses: -(2x(3x+3)), so:We get rid of parentheses
2x(3x+3)
We multiply parentheses
6x^2+6x
Back to the equation:
-(6x^2+6x)
3x^2-6x^2-3x+27x-6x-27=0
We add all the numbers together, and all the variables
-3x^2+18x-27=0
a = -3; b = 18; c = -27;
Δ = b2-4ac
Δ = 182-4·(-3)·(-27)
Δ = 0
Delta is equal to zero, so there is only one solution to the equation
Stosujemy wzór:$x=\frac{-b}{2a}=\frac{-18}{-6}=+3$
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