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

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

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

We add all the numbers together, and all the variables

-16t^2+t+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}
$