15=-16t^2+48t+3

Solution for 15=-16t^2+48t+3 equation:

15=-16t^2+48t+3

We move all terms to the left:

15-(-16t^2+48t+3)=0

We get rid of parentheses

16t^2-48t-3+15=0

We add all the numbers together, and all the variables

16t^2-48t+12=0

a = 16; b = -48; c = +12;
Δ = b2-4ac
Δ = -482-4·16·12
Δ = 1536
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:

$\sqrt{\Delta}=\sqrt{1536}=\sqrt{256*6}=\sqrt{256}*\sqrt{6}=16\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-16\sqrt{6}}{2*16}=\frac{48-16\sqrt{6}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+16\sqrt{6}}{2*16}=\frac{48+16\sqrt{6}}{32}
$