0=-16t^2+72+5

Solution for 0=-16t^2+72+5 equation:

0=-16t^2+72+5

We move all terms to the left:

0-(-16t^2+72+5)=0

We add all the numbers together, and all the variables

-(-16t^2+72+5)=0

We get rid of parentheses

16t^2-72-5=0

We add all the numbers together, and all the variables

16t^2-77=0

a = 16; b = 0; c = -77;
Δ = b2-4ac
Δ = 02-4·16·(-77)
Δ = 4928
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{4928}=\sqrt{64*77}=\sqrt{64}*\sqrt{77}=8\sqrt{77}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{77}}{2*16}=\frac{0-8\sqrt{77}}{32}
=-\frac{8\sqrt{77}}{32}
=-\frac{\sqrt{77}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{77}}{2*16}=\frac{0+8\sqrt{77}}{32}
=\frac{8\sqrt{77}}{32}
=\frac{\sqrt{77}}{4}
$