0=12.5t+-4.9t^2

Solution for 0=12.5t+-4.9t^2 equation:

0=12.5t+-4.9t^2

We move all terms to the left:

0-(12.5t+-4.9t^2)=0

We add all the numbers together, and all the variables

-(12.5t+-4.9t^2)=0

We use the square of the difference formula

-(12.5t-4.9t^2)=0

We get rid of parentheses

4.9t^2-12.5t=0

a = 4.9; b = -12.5; c = 0;
Δ = b2-4ac
Δ = -12.52-4·4.9·0
Δ = 156.25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12.5)-\sqrt{156.25}}{2*4.9}=\frac{12.5-\sqrt{156.25}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12.5)+\sqrt{156.25}}{2*4.9}=\frac{12.5+\sqrt{156.25}}{9.8}
$