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0=9x^2-3x
We move all terms to the left:
0-(9x^2-3x)=0
We add all the numbers together, and all the variables
-(9x^2-3x)=0
We get rid of parentheses
-9x^2+3x=0
a = -9; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-9)·0
Δ = 9
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{9}=3$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*-9}=\frac{-6}{-18} =1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-9}=\frac{0}{-18} =0 $
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