9p2+63p=0

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Solution for 9p2+63p=0 equation:



9p^2+63p=0
a = 9; b = 63; c = 0;
Δ = b2-4ac
Δ = 632-4·9·0
Δ = 3969
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3969}=63$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(63)-63}{2*9}=\frac{-126}{18} =-7 $
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(63)+63}{2*9}=\frac{0}{18} =0 $

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