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n^2+70n+275=0
a = 1; b = 70; c = +275;
Δ = b2-4ac
Δ = 702-4·1·275
Δ = 3800
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3800}=\sqrt{100*38}=\sqrt{100}*\sqrt{38}=10\sqrt{38}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(70)-10\sqrt{38}}{2*1}=\frac{-70-10\sqrt{38}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(70)+10\sqrt{38}}{2*1}=\frac{-70+10\sqrt{38}}{2} $
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