4y+12y^2=0

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Solution for 4y+12y^2=0 equation: 



4y+12y^2=0

a = 12; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·12·0
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*12}=\frac{-8}{24}
=-1/3
$


$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*12}=\frac{0}{24}
=0
$

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