6t^2-20t+8=0

Solution for 6t^2-20t+8=0 equation:

6t^2-20t+8=0

a = 6; b = -20; c = +8;
Δ = b2-4ac
Δ = -202-4·6·8
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-4\sqrt{13}}{2*6}=\frac{20-4\sqrt{13}}{12}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+4\sqrt{13}}{2*6}=\frac{20+4\sqrt{13}}{12}
$