16t^2-65t+40=0

Solution for 16t^2-65t+40=0 equation:

16t^2-65t+40=0

a = 16; b = -65; c = +40;
Δ = b2-4ac
Δ = -652-4·16·40
Δ = 1665
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:

$\sqrt{\Delta}=\sqrt{1665}=\sqrt{9*185}=\sqrt{9}*\sqrt{185}=3\sqrt{185}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-65)-3\sqrt{185}}{2*16}=\frac{65-3\sqrt{185}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-65)+3\sqrt{185}}{2*16}=\frac{65+3\sqrt{185}}{32}
$