0=16t+4t^2+0

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

0=16t+4t^2+0

We move all terms to the left:

0-(16t+4t^2+0)=0

We add all the numbers together, and all the variables

-(16t+4t^2+0)=0

We get rid of parentheses

-4t^2-16t-0=0

We add all the numbers together, and all the variables

-4t^2-16t=0

a = -4; b = -16; c = 0;
Δ = b2-4ac
Δ = -162-4·(-4)·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16}{2*-4}=\frac{0}{-8}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16}{2*-4}=\frac{32}{-8}
=-4
$