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3n^2+25n+42=0
a = 3; b = 25; c = +42;
Δ = b2-4ac
Δ = 252-4·3·42
Δ = 121
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{121}=11$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-11}{2*3}=\frac{-36}{6} =-6 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+11}{2*3}=\frac{-14}{6} =-2+1/3 $
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