16t^2+4t-0=0

Solution for 16t^2+4t-0=0 equation:

16t^2+4t-0=0

We add all the numbers together, and all the variables

16t^2+4t=0

a = 16; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·16·0
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*16}=\frac{-8}{32}
=-1/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*16}=\frac{0}{32}
=0
$