16t^2+9t=0

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Solution for 16t^2+9t=0 equation: 



16t^2+9t=0

a = 16; b = 9; c = 0;
Δ = b2-4ac
Δ = 92-4·16·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$


$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-9}{2*16}=\frac{-18}{32}
=-9/16
$


$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+9}{2*16}=\frac{0}{32}
=0
$

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