n^2+25n+100=0

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Solution for n^2+25n+100=0 equation: 



n^2+25n+100=0

a = 1; b = 25; c = +100;
Δ = b2-4ac
Δ = 252-4·1·100
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$


$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-15}{2*1}=\frac{-40}{2}
=-20
$


$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+15}{2*1}=\frac{-10}{2}
=-5
$

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