0=-49t^2+25t+.5

Solution for 0=-49t^2+25t+.5 equation:

0=-49t^2+25t+.5

We move all terms to the left:

0-(-49t^2+25t+.5)=0

We add all the numbers together, and all the variables

-(-49t^2+25t+.5)=0

We get rid of parentheses

49t^2-25t-.5=0

We add all the numbers together, and all the variables

49t^2-25t-0.5=0

a = 49; b = -25; c = -0.5;
Δ = b2-4ac
Δ = -252-4·49·(-0.5)
Δ = 723
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{723}}{2*49}=\frac{25-\sqrt{723}}{98}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{723}}{2*49}=\frac{25+\sqrt{723}}{98}
$