13=5+28t-16t^2

Solution for 13=5+28t-16t^2 equation:

13=5+28t-16t^2

We move all terms to the left:

13-(5+28t-16t^2)=0

We get rid of parentheses

16t^2-28t-5+13=0

We add all the numbers together, and all the variables

16t^2-28t+8=0

a = 16; b = -28; c = +8;
Δ = b2-4ac
Δ = -282-4·16·8
Δ = 272
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:

$\sqrt{\Delta}=\sqrt{272}=\sqrt{16*17}=\sqrt{16}*\sqrt{17}=4\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-4\sqrt{17}}{2*16}=\frac{28-4\sqrt{17}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4\sqrt{17}}{2*16}=\frac{28+4\sqrt{17}}{32}
$