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9n^2+45n+56=0
a = 9; b = 45; c = +56;
Δ = b2-4ac
Δ = 452-4·9·56
Δ = 9
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{9}=3$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(45)-3}{2*9}=\frac{-48}{18} =-2+2/3 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(45)+3}{2*9}=\frac{-42}{18} =-2+1/3 $
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