325=-16t^2+144t+5

Solution for 325=-16t^2+144t+5 equation:

325=-16t^2+144t+5

We move all terms to the left:

325-(-16t^2+144t+5)=0

We get rid of parentheses

16t^2-144t-5+325=0

We add all the numbers together, and all the variables

16t^2-144t+320=0

a = 16; b = -144; c = +320;
Δ = b2-4ac
Δ = -1442-4·16·320
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-144)-16}{2*16}=\frac{128}{32}
=4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-144)+16}{2*16}=\frac{160}{32}
=5
$