0=-16t^2-2t+203

Solution for 0=-16t^2-2t+203 equation:

0=-16t^2-2t+203

We move all terms to the left:

0-(-16t^2-2t+203)=0

We add all the numbers together, and all the variables

-(-16t^2-2t+203)=0

We get rid of parentheses

16t^2+2t-203=0

a = 16; b = 2; c = -203;
Δ = b2-4ac
Δ = 22-4·16·(-203)
Δ = 12996
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{12996}=114$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-114}{2*16}=\frac{-116}{32}
=-3+5/8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+114}{2*16}=\frac{112}{32}
=3+1/2
$