0=-16t^2+52t-13

Solution for 0=-16t^2+52t-13 equation:

0=-16t^2+52t-13

We move all terms to the left:

0-(-16t^2+52t-13)=0

We add all the numbers together, and all the variables

-(-16t^2+52t-13)=0

We get rid of parentheses

16t^2-52t+13=0

a = 16; b = -52; c = +13;
Δ = b2-4ac
Δ = -522-4·16·13
Δ = 1872
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:

$\sqrt{\Delta}=\sqrt{1872}=\sqrt{144*13}=\sqrt{144}*\sqrt{13}=12\sqrt{13}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-52)-12\sqrt{13}}{2*16}=\frac{52-12\sqrt{13}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-52)+12\sqrt{13}}{2*16}=\frac{52+12\sqrt{13}}{32}
$