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n^2+12n+10=0
a = 1; b = 12; c = +10;
Δ = b2-4ac
Δ = 122-4·1·10
Δ = 104
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{104}=\sqrt{4*26}=\sqrt{4}*\sqrt{26}=2\sqrt{26}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{26}}{2*1}=\frac{-12-2\sqrt{26}}{2} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{26}}{2*1}=\frac{-12+2\sqrt{26}}{2} $
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