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3c^2-2c-16=0
a = 3; b = -2; c = -16;
Δ = b2-4ac
Δ = -22-4·3·(-16)
Δ = 196
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}$$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{196}=14$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-14}{2*3}=\frac{-12}{6} =-2 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+14}{2*3}=\frac{16}{6} =2+2/3 $
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