0=-16t^2+t+49

Solution for 0=-16t^2+t+49 equation:

0=-16t^2+t+49

We move all terms to the left:

0-(-16t^2+t+49)=0

We add all the numbers together, and all the variables

-(-16t^2+t+49)=0

We get rid of parentheses

16t^2-t-49=0

We add all the numbers together, and all the variables

16t^2-1t-49=0

a = 16; b = -1; c = -49;
Δ = b2-4ac
Δ = -12-4·16·(-49)
Δ = 3137
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{3137}}{2*16}=\frac{1-\sqrt{3137}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{3137}}{2*16}=\frac{1+\sqrt{3137}}{32}
$