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3n^2+2n-1=-2n
We move all terms to the left:
3n^2+2n-1-(-2n)=0
We get rid of parentheses
3n^2+2n+2n-1=0
We add all the numbers together, and all the variables
3n^2+4n-1=0
a = 3; b = 4; c = -1;
Δ = b2-4ac
Δ = 42-4·3·(-1)
Δ = 28
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{7}}{2*3}=\frac{-4-2\sqrt{7}}{6} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{7}}{2*3}=\frac{-4+2\sqrt{7}}{6} $
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