0=-16t2+72t-45

Solution for 0=-16t2+72t-45 equation:

0=-16t^2+72t-45

We move all terms to the left:

0-(-16t^2+72t-45)=0

We add all the numbers together, and all the variables

-(-16t^2+72t-45)=0

We get rid of parentheses

16t^2-72t+45=0

a = 16; b = -72; c = +45;
Δ = b2-4ac
Δ = -722-4·16·45
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-72)-48}{2*16}=\frac{24}{32}
=3/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-72)+48}{2*16}=\frac{120}{32}
=3+3/4
$