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21n^2+41n+10=0
a = 21; b = 41; c = +10;
Δ = b2-4ac
Δ = 412-4·21·10
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$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-29}{2*21}=\frac{-70}{42} =-1+2/3 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+29}{2*21}=\frac{-12}{42} =-2/7 $
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