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3n^2=-n+2
We move all terms to the left:
3n^2-(-n+2)=0
We add all the numbers together, and all the variables
3n^2-(-1n+2)=0
We get rid of parentheses
3n^2+1n-2=0
We add all the numbers together, and all the variables
3n^2+n-2=0
a = 3; b = 1; c = -2;
Δ = b2-4ac
Δ = 12-4·3·(-2)
Δ = 25
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{25}=5$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-5}{2*3}=\frac{-6}{6} =-1 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+5}{2*3}=\frac{4}{6} =2/3 $
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