5p^2-17p+6=0

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Solution for 5p^2-17p+6=0 equation:



5p^2-17p+6=0
a = 5; b = -17; c = +6;
Δ = b2-4ac
Δ = -172-4·5·6
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-13}{2*5}=\frac{4}{10} =2/5 $
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+13}{2*5}=\frac{30}{10} =3 $

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