t=-16t^2+48t+160

Solution for t=-16t^2+48t+160 equation:

t=-16t^2+48t+160

We move all terms to the left:

t-(-16t^2+48t+160)=0

We get rid of parentheses

16t^2-48t+t-160=0

We add all the numbers together, and all the variables

16t^2-47t-160=0

a = 16; b = -47; c = -160;
Δ = b2-4ac
Δ = -472-4·16·(-160)
Δ = 12449
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{12449}}{2*16}=\frac{47-\sqrt{12449}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-47)+\sqrt{12449}}{2*16}=\frac{47+\sqrt{12449}}{32}
$