-4.9t^2+14.7t-11.02=0

Solution for -4.9t^2+14.7t-11.02=0 equation:

-4.9t^2+14.7t-11.02=0

a = -4.9; b = 14.7; c = -11.02;
Δ = b2-4ac
Δ = 14.72-4·(-4.9)·(-11.02)
Δ = 0.098
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{-(14.7)-\sqrt{0.098}}{2*-4.9}=\frac{-14.7-\sqrt{0.098}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14.7)+\sqrt{0.098}}{2*-4.9}=\frac{-14.7+\sqrt{0.098}}{-9.8}
$