1+25t-9.8t^2=0

Solution for 1+25t-9.8t^2=0 equation:

1+25t-9.8t^2=0

a = -9.8; b = 25; c = +1;
Δ = b2-4ac
Δ = 252-4·(-9.8)·1
Δ = 664.2
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{664.2}}{2*-9.8}=\frac{-25-\sqrt{664.2}}{-19.6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+\sqrt{664.2}}{2*-9.8}=\frac{-25+\sqrt{664.2}}{-19.6}
$