16t^2-305=0

Solution for 16t^2-305=0 equation:

16t^2-305=0

a = 16; b = 0; c = -305;
Δ = b2-4ac
Δ = 02-4·16·(-305)
Δ = 19520
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:

$\sqrt{\Delta}=\sqrt{19520}=\sqrt{64*305}=\sqrt{64}*\sqrt{305}=8\sqrt{305}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{305}}{2*16}=\frac{0-8\sqrt{305}}{32}
=-\frac{8\sqrt{305}}{32}
=-\frac{\sqrt{305}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{305}}{2*16}=\frac{0+8\sqrt{305}}{32}
=\frac{8\sqrt{305}}{32}
=\frac{\sqrt{305}}{4}
$