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2n^2-28n+48=0
a = 2; b = -28; c = +48;
Δ = b2-4ac
Δ = -282-4·2·48
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$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-20}{2*2}=\frac{8}{4} =2 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+20}{2*2}=\frac{48}{4} =12 $
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