3y^2-33y=0

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Solution for 3y^2-33y=0 equation: 



3y^2-33y=0

a = 3; b = -33; c = 0;
Δ = b2-4ac
Δ = -332-4·3·0
Δ = 1089
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1089}=33$


$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-33)-33}{2*3}=\frac{0}{6}
=0
$


$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+33}{2*3}=\frac{66}{6}
=11
$

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