8t^2-20t-3=0

Solution for 8t^2-20t-3=0 equation:

8t^2-20t-3=0

a = 8; b = -20; c = -3;
Δ = b2-4ac
Δ = -202-4·8·(-3)
Δ = 496
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:

$\sqrt{\Delta}=\sqrt{496}=\sqrt{16*31}=\sqrt{16}*\sqrt{31}=4\sqrt{31}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{31}}{2*8}=\frac{20-4\sqrt{31}}{16}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{31}}{2*8}=\frac{20+4\sqrt{31}}{16}
$