0=-2t^2+20t+22

Solution for 0=-2t^2+20t+22 equation:

0=-2t^2+20t+22

We move all terms to the left:

0-(-2t^2+20t+22)=0

We add all the numbers together, and all the variables

-(-2t^2+20t+22)=0

We get rid of parentheses

2t^2-20t-22=0

a = 2; b = -20; c = -22;
Δ = b2-4ac
Δ = -202-4·2·(-22)
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-24}{2*2}=\frac{-4}{4}
=-1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+24}{2*2}=\frac{44}{4}
=11
$