4.9t^2-11t+2.5=0

Solution for 4.9t^2-11t+2.5=0 equation:

4.9t^2-11t+2.5=0

a = 4.9; b = -11; c = +2.5;
Δ = b2-4ac
Δ = -112-4·4.9·2.5
Δ = 72
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-6\sqrt{2}}{2*4.9}=\frac{11-6\sqrt{2}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+6\sqrt{2}}{2*4.9}=\frac{11+6\sqrt{2}}{9.8}
$