0=16t^2+200t+25

Solution for 0=16t^2+200t+25 equation:

0=16t^2+200t+25

We move all terms to the left:

0-(16t^2+200t+25)=0

We add all the numbers together, and all the variables

-(16t^2+200t+25)=0

We get rid of parentheses

-16t^2-200t-25=0

a = -16; b = -200; c = -25;
Δ = b2-4ac
Δ = -2002-4·(-16)·(-25)
Δ = 38400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{38400}=\sqrt{6400*6}=\sqrt{6400}*\sqrt{6}=80\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-200)-80\sqrt{6}}{2*-16}=\frac{200-80\sqrt{6}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-200)+80\sqrt{6}}{2*-16}=\frac{200+80\sqrt{6}}{-32}
$