-16t^2+49t=0

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

-16t^2+49t=0

a = -16; b = 49; c = 0;
Δ = b2-4ac
Δ = 492-4·(-16)·0
Δ = 2401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2401}=49$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(49)-49}{2*-16}=\frac{-98}{-32}
=3+1/16
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(49)+49}{2*-16}=\frac{0}{-32}
=0
$