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3n^2+5n-288=0
a = 3; b = 5; c = -288;
Δ = b2-4ac
Δ = 52-4·3·(-288)
Δ = 3481
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{3481}=59$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-59}{2*3}=\frac{-64}{6} =-10+2/3 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+59}{2*3}=\frac{54}{6} =9 $
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