16t^2=205

Solution for 16t^2=205 equation:

16t^2=205

We move all terms to the left:

16t^2-(205)=0

a = 16; b = 0; c = -205;
Δ = b2-4ac
Δ = 02-4·16·(-205)
Δ = 13120
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:

$\sqrt{\Delta}=\sqrt{13120}=\sqrt{64*205}=\sqrt{64}*\sqrt{205}=8\sqrt{205}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{205}}{2*16}=\frac{0-8\sqrt{205}}{32}
=-\frac{8\sqrt{205}}{32}
=-\frac{\sqrt{205}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{205}}{2*16}=\frac{0+8\sqrt{205}}{32}
=\frac{8\sqrt{205}}{32}
=\frac{\sqrt{205}}{4}
$