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9=3x^2+6x
We move all terms to the left:
9-(3x^2+6x)=0
We get rid of parentheses
-3x^2-6x+9=0
a = -3; b = -6; c = +9;
Δ = b2-4ac
Δ = -62-4·(-3)·9
Δ = 144
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{144}=12$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-12}{2*-3}=\frac{-6}{-6} =1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+12}{2*-3}=\frac{18}{-6} =-3 $
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