-8t^2+39t+1=0

Solution for -8t^2+39t+1=0 equation:

-8t^2+39t+1=0

a = -8; b = 39; c = +1;
Δ = b2-4ac
Δ = 392-4·(-8)·1
Δ = 1553
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(39)-\sqrt{1553}}{2*-8}=\frac{-39-\sqrt{1553}}{-16}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+\sqrt{1553}}{2*-8}=\frac{-39+\sqrt{1553}}{-16}
$