-4.9t^2+1.5t+2.5=0

Solution for -4.9t^2+1.5t+2.5=0 equation:

-4.9t^2+1.5t+2.5=0

a = -4.9; b = 1.5; c = +2.5;
Δ = b2-4ac
Δ = 1.52-4·(-4.9)·2.5
Δ = 51.25
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{-(1.5)-\sqrt{51.25}}{2*-4.9}=\frac{-1.5-\sqrt{51.25}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.5)+\sqrt{51.25}}{2*-4.9}=\frac{-1.5+\sqrt{51.25}}{-9.8}
$