20=t^2-8t

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

20=t^2-8t

We move all terms to the left:

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

We get rid of parentheses

-t^2+8t+20=0

We add all the numbers together, and all the variables

-1t^2+8t+20=0

a = -1; b = 8; c = +20;
Δ = b2-4ac
Δ = 82-4·(-1)·20
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-12}{2*-1}=\frac{-20}{-2}
=+10
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+12}{2*-1}=\frac{4}{-2}
=-2
$