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3n^2+3n-1=5
We move all terms to the left:
3n^2+3n-1-(5)=0
We add all the numbers together, and all the variables
3n^2+3n-6=0
a = 3; b = 3; c = -6;
Δ = b2-4ac
Δ = 32-4·3·(-6)
Δ = 81
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{81}=9$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-9}{2*3}=\frac{-12}{6} =-2 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+9}{2*3}=\frac{6}{6} =1 $
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