-8t^2+32t-13=0

Solution for -8t^2+32t-13=0 equation:

-8t^2+32t-13=0

a = -8; b = 32; c = -13;
Δ = b2-4ac
Δ = 322-4·(-8)·(-13)
Δ = 608
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:

$\sqrt{\Delta}=\sqrt{608}=\sqrt{16*38}=\sqrt{16}*\sqrt{38}=4\sqrt{38}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-4\sqrt{38}}{2*-8}=\frac{-32-4\sqrt{38}}{-16}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+4\sqrt{38}}{2*-8}=\frac{-32+4\sqrt{38}}{-16}
$