0=10+20t-10t^2

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

0=10+20t-10t^2

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

10t^2-20t-10=0

a = 10; b = -20; c = -10;
Δ = b2-4ac
Δ = -202-4·10·(-10)
Δ = 800
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:

$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20\sqrt{2}}{2*10}=\frac{20-20\sqrt{2}}{20}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20\sqrt{2}}{2*10}=\frac{20+20\sqrt{2}}{20}
$