-9u^2+6u=0

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Solution for -9u^2+6u=0 equation: 



-9u^2+6u=0

a = -9; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·(-9)·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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


$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*-9}=\frac{-12}{-18}
=2/3
$


$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*-9}=\frac{0}{-18}
=0
$

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