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3n^2-21n+18=0
a = 3; b = -21; c = +18;
Δ = b2-4ac
Δ = -212-4·3·18
Δ = 225
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{225}=15$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-15}{2*3}=\frac{6}{6} =1 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+15}{2*3}=\frac{36}{6} =6 $
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