16t^2+300t+500=0

Solution for 16t^2+300t+500=0 equation:

16t^2+300t+500=0

a = 16; b = 300; c = +500;
Δ = b2-4ac
Δ = 3002-4·16·500
Δ = 58000
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:

$\sqrt{\Delta}=\sqrt{58000}=\sqrt{400*145}=\sqrt{400}*\sqrt{145}=20\sqrt{145}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(300)-20\sqrt{145}}{2*16}=\frac{-300-20\sqrt{145}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(300)+20\sqrt{145}}{2*16}=\frac{-300+20\sqrt{145}}{32}
$