7p^2+56p=0

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Solution for 7p^2+56p=0 equation: 



7p^2+56p=0

a = 7; b = 56; c = 0;
Δ = b2-4ac
Δ = 562-4·7·0
Δ = 3136
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3136}=56$


$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(56)-56}{2*7}=\frac{-112}{14}
=-8
$


$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(56)+56}{2*7}=\frac{0}{14}
=0
$

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