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3c^2-17c-14=0
a = 3; b = -17; c = -14;
Δ = b2-4ac
Δ = -172-4·3·(-14)
Δ = 457
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{-(-17)-\sqrt{457}}{2*3}=\frac{17-\sqrt{457}}{6} $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+\sqrt{457}}{2*3}=\frac{17+\sqrt{457}}{6} $
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