8=5-t^2+20t

Solution for 8=5-t^2+20t equation:

8=5-t^2+20t

We move all terms to the left:

8-(5-t^2+20t)=0

We get rid of parentheses

t^2-20t-5+8=0

We add all the numbers together, and all the variables

t^2-20t+3=0

a = 1; b = -20; c = +3;
Δ = b2-4ac
Δ = -202-4·1·3
Δ = 388
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:

$\sqrt{\Delta}=\sqrt{388}=\sqrt{4*97}=\sqrt{4}*\sqrt{97}=2\sqrt{97}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{97}}{2*1}=\frac{20-2\sqrt{97}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{97}}{2*1}=\frac{20+2\sqrt{97}}{2}
$