50+92t-16t^2=0

Solution for 50+92t-16t^2=0 equation:

50+92t-16t^2=0

a = -16; b = 92; c = +50;
Δ = b2-4ac
Δ = 922-4·(-16)·50
Δ = 11664
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{11664}=108$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(92)-108}{2*-16}=\frac{-200}{-32}
=6+1/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(92)+108}{2*-16}=\frac{16}{-32}
=-1/2
$