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4n^2+45n+11=0
a = 4; b = 45; c = +11;
Δ = b2-4ac
Δ = 452-4·4·11
Δ = 1849
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{1849}=43$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(45)-43}{2*4}=\frac{-88}{8} =-11 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(45)+43}{2*4}=\frac{-2}{8} =-1/4 $
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