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3c^2-5c-16=0
a = 3; b = -5; c = -16;
Δ = b2-4ac
Δ = -52-4·3·(-16)
Δ = 217
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{217}}{2*3}=\frac{5-\sqrt{217}}{6} $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{217}}{2*3}=\frac{5+\sqrt{217}}{6} $
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