4.905t^2-25t+15=0

Solution for 4.905t^2-25t+15=0 equation:

4.905t^2-25t+15=0

a = 4.905; b = -25; c = +15;
Δ = b2-4ac
Δ = -252-4·4.905·15
Δ = 330.7
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{330.7}}{2*4.905}=\frac{25-\sqrt{330.7}}{9.81}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{330.7}}{2*4.905}=\frac{25+\sqrt{330.7}}{9.81}
$