t^2-8t-4=0

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

t^2-8t-4=0

a = 1; b = -8; c = -4;
Δ = b2-4ac
Δ = -82-4·1·(-4)
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{5}}{2*1}=\frac{8-4\sqrt{5}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{5}}{2*1}=\frac{8+4\sqrt{5}}{2}
$