16t^2-300t+980=0

Solution for 16t^2-300t+980=0 equation:

16t^2-300t+980=0

a = 16; b = -300; c = +980;
Δ = b2-4ac
Δ = -3002-4·16·980
Δ = 27280
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:

$\sqrt{\Delta}=\sqrt{27280}=\sqrt{16*1705}=\sqrt{16}*\sqrt{1705}=4\sqrt{1705}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-300)-4\sqrt{1705}}{2*16}=\frac{300-4\sqrt{1705}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-300)+4\sqrt{1705}}{2*16}=\frac{300+4\sqrt{1705}}{32}
$