0=4.9t^2-25t+25

Solution for 0=4.9t^2-25t+25 equation:

0=4.9t^2-25t+25

We move all terms to the left:

0-(4.9t^2-25t+25)=0

We add all the numbers together, and all the variables

-(4.9t^2-25t+25)=0

We get rid of parentheses

-4.9t^2+25t-25=0

a = -4.9; b = 25; c = -25;
Δ = b2-4ac
Δ = 252-4·(-4.9)·(-25)
Δ = 135
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-\sqrt{135}}{2*-4.9}=\frac{-25-\sqrt{135}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+\sqrt{135}}{2*-4.9}=\frac{-25+\sqrt{135}}{-9.8}
$