0=t^2-8t-16

Solution for 0=t^2-8t-16 equation:

0=t^2-8t-16

We move all terms to the left:

0-(t^2-8t-16)=0

We add all the numbers together, and all the variables

-(t^2-8t-16)=0

We get rid of parentheses

-t^2+8t+16=0

We add all the numbers together, and all the variables

-1t^2+8t+16=0

a = -1; b = 8; c = +16;
Δ = b2-4ac
Δ = 82-4·(-1)·16
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{2}}{2*-1}=\frac{-8-8\sqrt{2}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{2}}{2*-1}=\frac{-8+8\sqrt{2}}{-2}
$