4t^2-16t-1=0

Solution for 4t^2-16t-1=0 equation:

4t^2-16t-1=0

a = 4; b = -16; c = -1;
Δ = b2-4ac
Δ = -162-4·4·(-1)
Δ = 272
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:

$\sqrt{\Delta}=\sqrt{272}=\sqrt{16*17}=\sqrt{16}*\sqrt{17}=4\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{17}}{2*4}=\frac{16-4\sqrt{17}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{17}}{2*4}=\frac{16+4\sqrt{17}}{8}
$