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3n^2+4n-32=0
a = 3; b = 4; c = -32;
Δ = b2-4ac
Δ = 42-4·3·(-32)
Δ = 400
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{400}=20$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-20}{2*3}=\frac{-24}{6} =-4 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+20}{2*3}=\frac{16}{6} =2+2/3 $
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