0=-16t^2+248t+13

Solution for 0=-16t^2+248t+13 equation:

0=-16t^2+248t+13

We move all terms to the left:

0-(-16t^2+248t+13)=0

We add all the numbers together, and all the variables

-(-16t^2+248t+13)=0

We get rid of parentheses

16t^2-248t-13=0

a = 16; b = -248; c = -13;
Δ = b2-4ac
Δ = -2482-4·16·(-13)
Δ = 62336
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:

$\sqrt{\Delta}=\sqrt{62336}=\sqrt{64*974}=\sqrt{64}*\sqrt{974}=8\sqrt{974}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-248)-8\sqrt{974}}{2*16}=\frac{248-8\sqrt{974}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-248)+8\sqrt{974}}{2*16}=\frac{248+8\sqrt{974}}{32}
$