-6t^2+64t-48=0

Solution for -6t^2+64t-48=0 equation:

-6t^2+64t-48=0

a = -6; b = 64; c = -48;
Δ = b2-4ac
Δ = 642-4·(-6)·(-48)
Δ = 2944
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:

$\sqrt{\Delta}=\sqrt{2944}=\sqrt{64*46}=\sqrt{64}*\sqrt{46}=8\sqrt{46}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-8\sqrt{46}}{2*-6}=\frac{-64-8\sqrt{46}}{-12}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+8\sqrt{46}}{2*-6}=\frac{-64+8\sqrt{46}}{-12}
$