t=-16t^2+48t+32

Solution for t=-16t^2+48t+32 equation:

t=-16t^2+48t+32

We move all terms to the left:

t-(-16t^2+48t+32)=0

We get rid of parentheses

16t^2-48t+t-32=0

We add all the numbers together, and all the variables

16t^2-47t-32=0

a = 16; b = -47; c = -32;
Δ = b2-4ac
Δ = -472-4·16·(-32)
Δ = 4257
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:

$\sqrt{\Delta}=\sqrt{4257}=\sqrt{9*473}=\sqrt{9}*\sqrt{473}=3\sqrt{473}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-47)-3\sqrt{473}}{2*16}=\frac{47-3\sqrt{473}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-47)+3\sqrt{473}}{2*16}=\frac{47+3\sqrt{473}}{32}
$