-4.9t^2+25t+3=0

Solution for -4.9t^2+25t+3=0 equation:

-4.9t^2+25t+3=0

a = -4.9; b = 25; c = +3;
Δ = b2-4ac
Δ = 252-4·(-4.9)·3
Δ = 683.8
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{683.8}}{2*-4.9}=\frac{-25-\sqrt{683.8}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+\sqrt{683.8}}{2*-4.9}=\frac{-25+\sqrt{683.8}}{-9.8}
$