4t^2-2=0

Solution for 4t^2-2=0 equation:

4t^2-2=0

a = 4; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·4·(-2)
Δ = 32
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:

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