0=4t^2-8t-4

Solution for 0=4t^2-8t-4 equation:

0=4t^2-8t-4

We move all terms to the left:

0-(4t^2-8t-4)=0

We add all the numbers together, and all the variables

-(4t^2-8t-4)=0

We get rid of parentheses

-4t^2+8t+4=0

a = -4; b = 8; c = +4;
Δ = b2-4ac
Δ = 82-4·(-4)·4
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*-4}=\frac{-8-8\sqrt{2}}{-8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{2}}{2*-4}=\frac{-8+8\sqrt{2}}{-8}
$