5p2-28p+32=0

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Solution for 5p2-28p+32=0 equation:



5p^2-28p+32=0
a = 5; b = -28; c = +32;
Δ = b2-4ac
Δ = -282-4·5·32
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-12}{2*5}=\frac{16}{10} =1+3/5 $
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+12}{2*5}=\frac{40}{10} =4 $

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