8t^2-48t+10=0

Solution for 8t^2-48t+10=0 equation:

8t^2-48t+10=0

a = 8; b = -48; c = +10;
Δ = b2-4ac
Δ = -482-4·8·10
Δ = 1984
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:

$\sqrt{\Delta}=\sqrt{1984}=\sqrt{64*31}=\sqrt{64}*\sqrt{31}=8\sqrt{31}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-8\sqrt{31}}{2*8}=\frac{48-8\sqrt{31}}{16}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+8\sqrt{31}}{2*8}=\frac{48+8\sqrt{31}}{16}
$