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.35n^2+22n+3=0
a = .35; b = 22; c = +3;
Δ = b2-4ac
Δ = 222-4·.35·3
Δ = 479.8
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-\sqrt{479.8}}{2*.35}=\frac{-22-\sqrt{479.8}}{0.7} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+\sqrt{479.8}}{2*.35}=\frac{-22+\sqrt{479.8}}{0.7} $
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