-3t^2+4t=0

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Solution for -3t^2+4t=0 equation: 



-3t^2+4t=0

a = -3; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-3)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$


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


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-3}=\frac{0}{-6}
=0
$

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