2+10t-4.9t^2=0

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

2+10t-4.9t^2=0

a = -4.9; b = 10; c = +2;
Δ = b2-4ac
Δ = 102-4·(-4.9)·2
Δ = 139.2
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{139.2}}{2*-4.9}=\frac{-10-\sqrt{139.2}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+\sqrt{139.2}}{2*-4.9}=\frac{-10+\sqrt{139.2}}{-9.8}
$