0=22t-5t^2+25

Solution for 0=22t-5t^2+25 equation:

0=22t-5t^2+25

We move all terms to the left:

0-(22t-5t^2+25)=0

We add all the numbers together, and all the variables

-(22t-5t^2+25)=0

We get rid of parentheses

5t^2-22t-25=0

a = 5; b = -22; c = -25;
Δ = b2-4ac
Δ = -222-4·5·(-25)
Δ = 984
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:

$\sqrt{\Delta}=\sqrt{984}=\sqrt{4*246}=\sqrt{4}*\sqrt{246}=2\sqrt{246}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{246}}{2*5}=\frac{22-2\sqrt{246}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{246}}{2*5}=\frac{22+2\sqrt{246}}{10}
$