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432=3n^2
We move all terms to the left:
432-(3n^2)=0
a = -3; b = 0; c = +432;
Δ = b2-4ac
Δ = 02-4·(-3)·432
Δ = 5184
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{5184}=72$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-72}{2*-3}=\frac{-72}{-6} =+12 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+72}{2*-3}=\frac{72}{-6} =-12 $
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