0=2(t^2)+5t

Solution for 0=2(t^2)+5t equation:

0=2(t^2)+5t

We move all terms to the left:

0-(2(t^2)+5t)=0

We add all the numbers together, and all the variables

-(2t^2+5t)=0

We get rid of parentheses

-2t^2-5t=0

a = -2; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·(-2)·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*-2}=\frac{0}{-4}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*-2}=\frac{10}{-4}
=-2+1/2
$