10=-16t^2+48t+8

Solution for 10=-16t^2+48t+8 equation:

10=-16t^2+48t+8

We move all terms to the left:

10-(-16t^2+48t+8)=0

We get rid of parentheses

16t^2-48t-8+10=0

We add all the numbers together, and all the variables

16t^2-48t+2=0

a = 16; b = -48; c = +2;
Δ = b2-4ac
Δ = -482-4·16·2
Δ = 2176
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:

$\sqrt{\Delta}=\sqrt{2176}=\sqrt{64*34}=\sqrt{64}*\sqrt{34}=8\sqrt{34}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-8\sqrt{34}}{2*16}=\frac{48-8\sqrt{34}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+8\sqrt{34}}{2*16}=\frac{48+8\sqrt{34}}{32}
$