16t^2+320t+50=0

Solution for 16t^2+320t+50=0 equation:

16t^2+320t+50=0

a = 16; b = 320; c = +50;
Δ = b2-4ac
Δ = 3202-4·16·50
Δ = 99200
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:

$\sqrt{\Delta}=\sqrt{99200}=\sqrt{1600*62}=\sqrt{1600}*\sqrt{62}=40\sqrt{62}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(320)-40\sqrt{62}}{2*16}=\frac{-320-40\sqrt{62}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(320)+40\sqrt{62}}{2*16}=\frac{-320+40\sqrt{62}}{32}
$