p^2-20p+99=0

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Solution for p^2-20p+99=0 equation: 



p^2-20p+99=0

a = 1; b = -20; c = +99;
Δ = b2-4ac
Δ = -202-4·1·99
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$


$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2}{2*1}=\frac{18}{2}
=9
$


$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2}{2*1}=\frac{22}{2}
=11
$

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