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3n^2+2n-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:$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{196}=14$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-14}{2*3}=\frac{-16}{6} =-2+2/3 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+14}{2*3}=\frac{12}{6} =2 $
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