-6y+3y^2=0

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Solution for -6y+3y^2=0 equation: 



-6y+3y^2=0

a = 3; b = -6; c = 0;
Δ = b2-4ac
Δ = -62-4·3·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$


$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6}{2*3}=\frac{0}{6}
=0
$


$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6}{2*3}=\frac{12}{6}
=2
$

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