(2^x)*4^(x^2)=2

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

(2^x)*4^(x^2)=2

We move all terms to the left:

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

Wy multiply elements

8x^2-2=0

a = 8; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·8·(-2)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8}{2*8}=\frac{-8}{16}
=-1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*8}=\frac{8}{16}
=1/2
$