0=-16t^2+16t+52

Solution for 0=-16t^2+16t+52 equation:

0=-16t^2+16t+52

We move all terms to the left:

0-(-16t^2+16t+52)=0

We add all the numbers together, and all the variables

-(-16t^2+16t+52)=0

We get rid of parentheses

16t^2-16t-52=0

a = 16; b = -16; c = -52;
Δ = b2-4ac
Δ = -162-4·16·(-52)
Δ = 3584
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:

$\sqrt{\Delta}=\sqrt{3584}=\sqrt{256*14}=\sqrt{256}*\sqrt{14}=16\sqrt{14}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16\sqrt{14}}{2*16}=\frac{16-16\sqrt{14}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16\sqrt{14}}{2*16}=\frac{16+16\sqrt{14}}{32}
$