3y^2+60y=0

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Solution for 3y^2+60y=0 equation: 



3y^2+60y=0

a = 3; b = 60; c = 0;
Δ = b2-4ac
Δ = 602-4·3·0
Δ = 3600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3600}=60$


$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-60}{2*3}=\frac{-120}{6}
=-20
$


$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+60}{2*3}=\frac{0}{6}
=0
$

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