4t^2-20t-1=0

Solution for 4t^2-20t-1=0 equation:

4t^2-20t-1=0

a = 4; b = -20; c = -1;
Δ = b2-4ac
Δ = -202-4·4·(-1)
Δ = 416
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:

$\sqrt{\Delta}=\sqrt{416}=\sqrt{16*26}=\sqrt{16}*\sqrt{26}=4\sqrt{26}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{26}}{2*4}=\frac{20-4\sqrt{26}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{26}}{2*4}=\frac{20+4\sqrt{26}}{8}
$