45=-16t^2+96t+t

Solution for 45=-16t^2+96t+t equation:

45=-16t^2+96t+t

We move all terms to the left:

45-(-16t^2+96t+t)=0

We get rid of parentheses

16t^2-96t-t+45=0

We add all the numbers together, and all the variables

16t^2-97t+45=0

a = 16; b = -97; c = +45;
Δ = b2-4ac
Δ = -972-4·16·45
Δ = 6529
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{-(-97)-\sqrt{6529}}{2*16}=\frac{97-\sqrt{6529}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-97)+\sqrt{6529}}{2*16}=\frac{97+\sqrt{6529}}{32}
$