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12=n(n-3)
We move all terms to the left:
12-(n(n-3))=0
We calculate terms in parentheses: -(n(n-3)), so:We get rid of parentheses
n(n-3)
We multiply parentheses
n^2-3n
Back to the equation:
-(n^2-3n)
-n^2+3n+12=0
We add all the numbers together, and all the variables
-1n^2+3n+12=0
a = -1; b = 3; c = +12;
Δ = b2-4ac
Δ = 32-4·(-1)·12
Δ = 57
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{57}}{2*-1}=\frac{-3-\sqrt{57}}{-2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{57}}{2*-1}=\frac{-3+\sqrt{57}}{-2} $
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