0.25t^2+16t=0

Solution for 0.25t^2+16t=0 equation:

0.25t^2+16t=0

a = 0.25; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·0.25·0
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*0.25}=\frac{-32}{0.5}
=-64
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*0.25}=\frac{0}{0.5}
=0
$