(28y^2)+5y=0

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Solution for (28y^2)+5y=0 equation: 



(28y^2)+5y=0

a = 28; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·28·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$


$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*28}=\frac{-10}{56}
=-5/28
$


$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*28}=\frac{0}{56}
=0
$

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