-3c^2+8c=0

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Solution for -3c^2+8c=0 equation: 



-3c^2+8c=0

a = -3; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·(-3)·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$


$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*-3}=\frac{-16}{-6}
=2+2/3
$


$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*-3}=\frac{0}{-6}
=0
$

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