16t^2-300t+600=0

Solution for 16t^2-300t+600=0 equation:

16t^2-300t+600=0

a = 16; b = -300; c = +600;
Δ = b2-4ac
Δ = -3002-4·16·600
Δ = 51600
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:

$\sqrt{\Delta}=\sqrt{51600}=\sqrt{400*129}=\sqrt{400}*\sqrt{129}=20\sqrt{129}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-300)-20\sqrt{129}}{2*16}=\frac{300-20\sqrt{129}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-300)+20\sqrt{129}}{2*16}=\frac{300+20\sqrt{129}}{32}
$