16t^2+200t+350=0

Solution for 16t^2+200t+350=0 equation:

16t^2+200t+350=0

a = 16; b = 200; c = +350;
Δ = b2-4ac
Δ = 2002-4·16·350
Δ = 17600
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:

$\sqrt{\Delta}=\sqrt{17600}=\sqrt{1600*11}=\sqrt{1600}*\sqrt{11}=40\sqrt{11}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(200)-40\sqrt{11}}{2*16}=\frac{-200-40\sqrt{11}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(200)+40\sqrt{11}}{2*16}=\frac{-200+40\sqrt{11}}{32}
$