0=10+20t-5t^2

Solution for 0=10+20t-5t^2 equation:

0=10+20t-5t^2

We move all terms to the left:

0-(10+20t-5t^2)=0

We add all the numbers together, and all the variables

-(10+20t-5t^2)=0

We get rid of parentheses

5t^2-20t-10=0

a = 5; b = -20; c = -10;
Δ = b2-4ac
Δ = -202-4·5·(-10)
Δ = 600
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:

$\sqrt{\Delta}=\sqrt{600}=\sqrt{100*6}=\sqrt{100}*\sqrt{6}=10\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-10\sqrt{6}}{2*5}=\frac{20-10\sqrt{6}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+10\sqrt{6}}{2*5}=\frac{20+10\sqrt{6}}{10}
$