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3n^2+11n-20=0
a = 3; b = 11; c = -20;
Δ = b2-4ac
Δ = 112-4·3·(-20)
Δ = 361
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{361}=19$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-19}{2*3}=\frac{-30}{6} =-5 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+19}{2*3}=\frac{8}{6} =1+1/3 $
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