272=-16t^2+136t+14

Solution for 272=-16t^2+136t+14 equation:

272=-16t^2+136t+14

We move all terms to the left:

272-(-16t^2+136t+14)=0

We get rid of parentheses

16t^2-136t-14+272=0

We add all the numbers together, and all the variables

16t^2-136t+258=0

a = 16; b = -136; c = +258;
Δ = b2-4ac
Δ = -1362-4·16·258
Δ = 1984
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:

$\sqrt{\Delta}=\sqrt{1984}=\sqrt{64*31}=\sqrt{64}*\sqrt{31}=8\sqrt{31}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-136)-8\sqrt{31}}{2*16}=\frac{136-8\sqrt{31}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-136)+8\sqrt{31}}{2*16}=\frac{136+8\sqrt{31}}{32}
$