p^2-20p+84=0

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Solution for p^2-20p+84=0 equation: 



p^2-20p+84=0

a = 1; b = -20; c = +84;
Δ = b2-4ac
Δ = -202-4·1·84
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$


$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-8}{2*1}=\frac{12}{2}
=6
$


$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+8}{2*1}=\frac{28}{2}
=14
$

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