13h^2+27h=0

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Solution for 13h^2+27h=0 equation: 



13h^2+27h=0

a = 13; b = 27; c = 0;
Δ = b2-4ac
Δ = 272-4·13·0
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$


$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-27}{2*13}=\frac{-54}{26}
=-2+1/13
$


$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+27}{2*13}=\frac{0}{26}
=0
$

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