-8t^2-2t=0

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Solution for -8t^2-2t=0 equation: 



-8t^2-2t=0

a = -8; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·(-8)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*-8}=\frac{0}{-16}
=0
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*-8}=\frac{4}{-16}
=-1/4
$

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