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2n^2-19n+42=0
a = 2; b = -19; c = +42;
Δ = b2-4ac
Δ = -192-4·2·42
Δ = 25
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{25}=5$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-5}{2*2}=\frac{14}{4} =3+1/2 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+5}{2*2}=\frac{24}{4} =6 $
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