0=4.9t^2-20t+10

Solution for 0=4.9t^2-20t+10 equation:

0=4.9t^2-20t+10

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

a = -4.9; b = 20; c = -10;
Δ = b2-4ac
Δ = 202-4·(-4.9)·(-10)
Δ = 204
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{204}=\sqrt{4*51}=\sqrt{4}*\sqrt{51}=2\sqrt{51}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{51}}{2*-4.9}=\frac{-20-2\sqrt{51}}{-9.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{51}}{2*-4.9}=\frac{-20+2\sqrt{51}}{-9.8}
$