0=-2t^2+20t+8

Solution for 0=-2t^2+20t+8 equation:

0=-2t^2+20t+8

We move all terms to the left:

0-(-2t^2+20t+8)=0

We add all the numbers together, and all the variables

-(-2t^2+20t+8)=0

We get rid of parentheses

2t^2-20t-8=0

a = 2; b = -20; c = -8;
Δ = b2-4ac
Δ = -202-4·2·(-8)
Δ = 464
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:

$\sqrt{\Delta}=\sqrt{464}=\sqrt{16*29}=\sqrt{16}*\sqrt{29}=4\sqrt{29}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{29}}{2*2}=\frac{20-4\sqrt{29}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{29}}{2*2}=\frac{20+4\sqrt{29}}{4}
$