0=248(t)-16t^2

Solution for 0=248(t)-16t^2 equation:

0=248(t)-16t^2

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

16t^2-248t=0

a = 16; b = -248; c = 0;
Δ = b2-4ac
Δ = -2482-4·16·0
Δ = 61504
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{61504}=248$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-248)-248}{2*16}=\frac{0}{32}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-248)+248}{2*16}=\frac{496}{32}
=15+1/2
$