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3x^2=9x=0

We move all terms to the left:

3x^2-(9x)=0

a = 3; b = -9; c = 0;

Δ = b^{2}-4ac

Δ = -9^{2}-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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-9}{2*3}=\frac{0}{6} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+9}{2*3}=\frac{18}{6} =3 $

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