4t^2-128=0

Solution for 4t^2-128=0 equation:

4t^2-128=0

a = 4; b = 0; c = -128;
Δ = b2-4ac
Δ = 02-4·4·(-128)
Δ = 2048
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:

$\sqrt{\Delta}=\sqrt{2048}=\sqrt{1024*2}=\sqrt{1024}*\sqrt{2}=32\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32\sqrt{2}}{2*4}=\frac{0-32\sqrt{2}}{8}
=-\frac{32\sqrt{2}}{8}
=-4\sqrt{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32\sqrt{2}}{2*4}=\frac{0+32\sqrt{2}}{8}
=\frac{32\sqrt{2}}{8}
=4\sqrt{2}
$