16t^2+54t+16=0

Solution for 16t^2+54t+16=0 equation:

16t^2+54t+16=0

a = 16; b = 54; c = +16;
Δ = b2-4ac
Δ = 542-4·16·16
Δ = 1892
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:

$\sqrt{\Delta}=\sqrt{1892}=\sqrt{4*473}=\sqrt{4}*\sqrt{473}=2\sqrt{473}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(54)-2\sqrt{473}}{2*16}=\frac{-54-2\sqrt{473}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(54)+2\sqrt{473}}{2*16}=\frac{-54+2\sqrt{473}}{32}
$