-5t^2+6t=0

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Solution for -5t^2+6t=0 equation: 



-5t^2+6t=0

a = -5; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·(-5)·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*-5}=\frac{-12}{-10}
=1+1/5
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*-5}=\frac{0}{-10}
=0
$

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