0=-104+22t+t^2

Solution for 0=-104+22t+t^2 equation:

0=-104+22t+t^2

We move all terms to the left:

0-(-104+22t+t^2)=0

We add all the numbers together, and all the variables

-(-104+22t+t^2)=0

We get rid of parentheses

-t^2-22t+104=0

We add all the numbers together, and all the variables

-1t^2-22t+104=0

a = -1; b = -22; c = +104;
Δ = b2-4ac
Δ = -222-4·(-1)·104
Δ = 900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{900}=30$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-30}{2*-1}=\frac{-8}{-2}
=+4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+30}{2*-1}=\frac{52}{-2}
=-26
$