-8t^2+64t+6=0

Solution for -8t^2+64t+6=0 equation:

-8t^2+64t+6=0

a = -8; b = 64; c = +6;
Δ = b2-4ac
Δ = 642-4·(-8)·6
Δ = 4288
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:

$\sqrt{\Delta}=\sqrt{4288}=\sqrt{64*67}=\sqrt{64}*\sqrt{67}=8\sqrt{67}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-8\sqrt{67}}{2*-8}=\frac{-64-8\sqrt{67}}{-16}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+8\sqrt{67}}{2*-8}=\frac{-64+8\sqrt{67}}{-16}
$