164t-16t^2=0

Solution for 164t-16t^2=0 equation:

164t-16t^2=0

a = -16; b = 164; c = 0;
Δ = b2-4ac
Δ = 1642-4·(-16)·0
Δ = 26896
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{26896}=164$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(164)-164}{2*-16}=\frac{-328}{-32}
=10+1/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(164)+164}{2*-16}=\frac{0}{-32}
=0
$