0=t^2-20t-64

Solution for 0=t^2-20t-64 equation:

0=t^2-20t-64

We move all terms to the left:

0-(t^2-20t-64)=0

We add all the numbers together, and all the variables

-(t^2-20t-64)=0

We get rid of parentheses

-t^2+20t+64=0

We add all the numbers together, and all the variables

-1t^2+20t+64=0

a = -1; b = 20; c = +64;
Δ = b2-4ac
Δ = 202-4·(-1)·64
Δ = 656
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:

$\sqrt{\Delta}=\sqrt{656}=\sqrt{16*41}=\sqrt{16}*\sqrt{41}=4\sqrt{41}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{41}}{2*-1}=\frac{-20-4\sqrt{41}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{41}}{2*-1}=\frac{-20+4\sqrt{41}}{-2}
$