10=t+4.9t^2

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

10=t+4.9t^2

We move all terms to the left:

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

We get rid of parentheses

-4.9t^2-t+10=0

We add all the numbers together, and all the variables

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

a = -4.9; b = -1; c = +10;
Δ = b2-4ac
Δ = -12-4·(-4.9)·10
Δ = 197
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{-(-1)-\sqrt{197}}{2*-4.9}=\frac{1-\sqrt{197}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{197}}{2*-4.9}=\frac{1+\sqrt{197}}{-9.8}
$