0=-6(t^2)+20t+800

Solution for 0=-6(t^2)+20t+800 equation:

0=-6(t^2)+20t+800

We move all terms to the left:

0-(-6(t^2)+20t+800)=0

We add all the numbers together, and all the variables

-(-6t^2+20t+800)=0

We get rid of parentheses

6t^2-20t-800=0

a = 6; b = -20; c = -800;
Δ = b2-4ac
Δ = -202-4·6·(-800)
Δ = 19600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{19600}=140$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-140}{2*6}=\frac{-120}{12}
=-10
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+140}{2*6}=\frac{160}{12}
=13+1/3
$