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3n^2-9n+6=0
a = 3; b = -9; c = +6;
Δ = b2-4ac
Δ = -92-4·3·6
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{9}=3$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-3}{2*3}=\frac{6}{6} =1 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+3}{2*3}=\frac{12}{6} =2 $
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