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3n^2-19n-40=0
a = 3; b = -19; c = -40;
Δ = b2-4ac
Δ = -192-4·3·(-40)
Δ = 841
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{841}=29$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-29}{2*3}=\frac{-10}{6} =-1+2/3 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+29}{2*3}=\frac{48}{6} =8 $
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