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15n^2+32n+16=0
a = 15; b = 32; c = +16;
Δ = b2-4ac
Δ = 322-4·15·16
Δ = 64
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{64}=8$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-8}{2*15}=\frac{-40}{30} =-1+1/3 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+8}{2*15}=\frac{-24}{30} =-4/5 $
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