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48n^2+18n-30=0
a = 48; b = 18; c = -30;
Δ = b2-4ac
Δ = 182-4·48·(-30)
Δ = 6084
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{6084}=78$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-78}{2*48}=\frac{-96}{96} =-1 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+78}{2*48}=\frac{60}{96} =5/8 $
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