-16t^2+152t=0

Solution for -16t^2+152t=0 equation:

-16t^2+152t=0

a = -16; b = 152; c = 0;
Δ = b2-4ac
Δ = 1522-4·(-16)·0
Δ = 23104
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{23104}=152$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(152)-152}{2*-16}=\frac{-304}{-32}
=9+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(152)+152}{2*-16}=\frac{0}{-32}
=0
$