2+10t-4.9t^2=1

Solution for 2+10t-4.9t^2=1 equation:

2+10t-4.9t^2=1

We move all terms to the left:

2+10t-4.9t^2-(1)=0

We add all the numbers together, and all the variables

-4.9t^2+10t+1=0

a = -4.9; b = 10; c = +1;
Δ = b2-4ac
Δ = 102-4·(-4.9)·1
Δ = 119.6
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{119.6}}{2*-4.9}=\frac{-10-\sqrt{119.6}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+\sqrt{119.6}}{2*-4.9}=\frac{-10+\sqrt{119.6}}{-9.8}
$