-4.9t^2+10t-5.1=0

Solution for -4.9t^2+10t-5.1=0 equation:

-4.9t^2+10t-5.1=0

a = -4.9; b = 10; c = -5.1;
Δ = b2-4ac
Δ = 102-4·(-4.9)·(-5.1)
Δ = 0.040000000000006
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{-(10)-\sqrt{0.040000000000006}}{2*-4.9}=\frac{-10-\sqrt{0.040000000000006}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+\sqrt{0.040000000000006}}{2*-4.9}=\frac{-10+\sqrt{0.040000000000006}}{-9.8}
$