25*t-5*t^2=0

Solution for 25*t-5*t^2=0 equation:

25t-5t^2=0

a = -5; b = 25; c = 0;
Δ = b2-4ac
Δ = 252-4·(-5)·0
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-25}{2*-5}=\frac{-50}{-10}
=+5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+25}{2*-5}=\frac{0}{-10}
=0
$