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16n^2-48n=0
a = 16; b = -48; c = 0;
Δ = b2-4ac
Δ = -482-4·16·0
Δ = 2304
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{2304}=48$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-48}{2*16}=\frac{0}{32} =0 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+48}{2*16}=\frac{96}{32} =3 $
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