t=-16t2+48t+4

Solution for t=-16t2+48t+4 equation:

t=-16t^2+48t+4

We move all terms to the left:

t-(-16t^2+48t+4)=0

We get rid of parentheses

16t^2-48t+t-4=0

We add all the numbers together, and all the variables

16t^2-47t-4=0

a = 16; b = -47; c = -4;
Δ = b2-4ac
Δ = -472-4·16·(-4)
Δ = 2465
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-47)-\sqrt{2465}}{2*16}=\frac{47-\sqrt{2465}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-47)+\sqrt{2465}}{2*16}=\frac{47+\sqrt{2465}}{32}
$