(y)^2+4y=0

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Solution for (y)^2+4y=0 equation: 



(y)^2+4y=0

a = 1; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·1·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$


$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*1}=\frac{-8}{2}
=-4
$


$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*1}=\frac{0}{2}
=0
$

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