16t^2+600t=0

Solution for 16t^2+600t=0 equation:

16t^2+600t=0

a = 16; b = 600; c = 0;
Δ = b2-4ac
Δ = 6002-4·16·0
Δ = 360000
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{360000}=600$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(600)-600}{2*16}=\frac{-1200}{32}
=-37+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(600)+600}{2*16}=\frac{0}{32}
=0
$