64^(2x)*4=32

Solution for 64^(2x)*4=32 equation:

64^(2x)*4=32

We move all terms to the left:

64^(2x)*4-(32)=0

Wy multiply elements

256x^2-32=0

a = 256; b = 0; c = -32;
Δ = b2-4ac
Δ = 02-4·256·(-32)
Δ = 32768
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{32768}=\sqrt{16384*2}=\sqrt{16384}*\sqrt{2}=128\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-128\sqrt{2}}{2*256}=\frac{0-128\sqrt{2}}{512}
=-\frac{128\sqrt{2}}{512}
=-\frac{\sqrt{2}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+128\sqrt{2}}{2*256}=\frac{0+128\sqrt{2}}{512}
=\frac{128\sqrt{2}}{512}
=\frac{\sqrt{2}}{4}
$