4t^2-8t-4=0

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

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}
$