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10n^2+27n+12=0
a = 10; b = 27; c = +12;
Δ = b2-4ac
Δ = 272-4·10·12
Δ = 249
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{-(27)-\sqrt{249}}{2*10}=\frac{-27-\sqrt{249}}{20} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+\sqrt{249}}{2*10}=\frac{-27+\sqrt{249}}{20} $
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