10=-5t^2+20t+2

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

10=-5t^2+20t+2

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

5t^2-20t+8=0

a = 5; b = -20; c = +8;
Δ = b2-4ac
Δ = -202-4·5·8
Δ = 240
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:

$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{15}}{2*5}=\frac{20-4\sqrt{15}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{15}}{2*5}=\frac{20+4\sqrt{15}}{10}
$