0=-16t^2+48t+19

Solution for 0=-16t^2+48t+19 equation:

0=-16t^2+48t+19

We move all terms to the left:

0-(-16t^2+48t+19)=0

We add all the numbers together, and all the variables

-(-16t^2+48t+19)=0

We get rid of parentheses

16t^2-48t-19=0

a = 16; b = -48; c = -19;
Δ = b2-4ac
Δ = -482-4·16·(-19)
Δ = 3520
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:

$\sqrt{\Delta}=\sqrt{3520}=\sqrt{64*55}=\sqrt{64}*\sqrt{55}=8\sqrt{55}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-8\sqrt{55}}{2*16}=\frac{48-8\sqrt{55}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+8\sqrt{55}}{2*16}=\frac{48+8\sqrt{55}}{32}
$