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n^2+1n=210
We move all terms to the left:
n^2+1n-(210)=0
We add all the numbers together, and all the variables
n^2+n-210=0
a = 1; b = 1; c = -210;
Δ = b2-4ac
Δ = 12-4·1·(-210)
Δ = 841
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{841}=29$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-29}{2*1}=\frac{-30}{2} =-15 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+29}{2*1}=\frac{28}{2} =14 $
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