t=-4.9t^2+20t+10

Solution for t=-4.9t^2+20t+10 equation:

t=-4.9t^2+20t+10

We move all terms to the left:

t-(-4.9t^2+20t+10)=0

We get rid of parentheses

4.9t^2-20t+t-10=0

We add all the numbers together, and all the variables

4.9t^2-19t-10=0

a = 4.9; b = -19; c = -10;
Δ = b2-4ac
Δ = -192-4·4.9·(-10)
Δ = 557
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{-(-19)-\sqrt{557}}{2*4.9}=\frac{19-\sqrt{557}}{9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{557}}{2*4.9}=\frac{19+\sqrt{557}}{9.8}
$