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(3n-1)n=4
We move all terms to the left:
(3n-1)n-(4)=0
We multiply parentheses
3n^2-1n-4=0
a = 3; b = -1; c = -4;
Δ = b2-4ac
Δ = -12-4·3·(-4)
Δ = 49
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{49}=7$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-7}{2*3}=\frac{-6}{6} =-1 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+7}{2*3}=\frac{8}{6} =1+1/3 $
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