-16t2+50t+3.5=0

Solution for -16t2+50t+3.5=0 equation:

-16t^2+50t+3.5=0

a = -16; b = 50; c = +3.5;
Δ = b2-4ac
Δ = 502-4·(-16)·3.5
Δ = 2724
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2724}=\sqrt{4*681}=\sqrt{4}*\sqrt{681}=2\sqrt{681}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-2\sqrt{681}}{2*-16}=\frac{-50-2\sqrt{681}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+2\sqrt{681}}{2*-16}=\frac{-50+2\sqrt{681}}{-32}
$