3y+4y^2=0

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Solution for 3y+4y^2=0 equation: 



3y+4y^2=0

a = 4; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·4·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$


$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*4}=\frac{-6}{8}
=-3/4
$


$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*4}=\frac{0}{8}
=0
$

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