25t+.5t^2=0

Solution for 25t+.5t^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}{1}
=-50
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+25}{2*.5}=\frac{0}{1}
=0
$