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3d^2-16d+5=0
a = 3; b = -16; c = +5;
Δ = b2-4ac
Δ = -162-4·3·5
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{196}=14$$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-14}{2*3}=\frac{2}{6} =1/3 $$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+14}{2*3}=\frac{30}{6} =5 $
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