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3n^2-15n-550=0
a = 3; b = -15; c = -550;
Δ = b2-4ac
Δ = -152-4·3·(-550)
Δ = 6825
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{6825}=\sqrt{25*273}=\sqrt{25}*\sqrt{273}=5\sqrt{273}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-5\sqrt{273}}{2*3}=\frac{15-5\sqrt{273}}{6} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+5\sqrt{273}}{2*3}=\frac{15+5\sqrt{273}}{6} $
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