1=-16t^2+47t+0

Solution for 1=-16t^2+47t+0 equation:

1=-16t^2+47t+0

We move all terms to the left:

1-(-16t^2+47t+0)=0

We get rid of parentheses

16t^2-47t-0+1=0

We add all the numbers together, and all the variables

16t^2-47t+1=0

a = 16; b = -47; c = +1;
Δ = b2-4ac
Δ = -472-4·16·1
Δ = 2145
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{2145}}{2*16}=\frac{47-\sqrt{2145}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-47)+\sqrt{2145}}{2*16}=\frac{47+\sqrt{2145}}{32}
$