3n2=-n+14

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Solution for 3n2=-n+14 equation:



3n^2=-n+14
We move all terms to the left:
3n^2-(-n+14)=0
We add all the numbers together, and all the variables
3n^2-(-1n+14)=0
We get rid of parentheses
3n^2+1n-14=0
We add all the numbers together, and all the variables
3n^2+n-14=0
a = 3; b = 1; c = -14;
Δ = b2-4ac
Δ = 12-4·3·(-14)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-13}{2*3}=\frac{-14}{6} =-2+1/3 $
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+13}{2*3}=\frac{12}{6} =2 $

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