4y2+13y=0

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Solution for 4y2+13y=0 equation: 



4y^2+13y=0

a = 4; b = 13; c = 0;
Δ = b2-4ac
Δ = 132-4·4·0
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$


$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-13}{2*4}=\frac{-26}{8}
=-3+1/4
$


$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+13}{2*4}=\frac{0}{8}
=0
$

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