16t2=4

Solution for 16t2=4 equation:

16t^2=4

We move all terms to the left:

16t^2-(4)=0

a = 16; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·16·(-4)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*16}=\frac{-16}{32}
=-1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*16}=\frac{16}{32}
=1/2
$