(x^2)+(7(3^-2))=(2x)^2

Solution for (x^2)+(7(3^-2))=(2x)^2 equation:

(x^2)+(7(3^-2))=(2x)^2

We move all terms to the left:

(x^2)+(7(3^-2))-((2x)^2)=0

determiningTheFunctionDomain
x^2-2x^2+(7(3^-2))=0

We add all the numbers together, and all the variables

-1x^2+0.77777777777778=0

a = -1; b = 0; c = +0.77777777777778;
Δ = b2-4ac
Δ = 02-4·(-1)·0.77777777777778
Δ = 3.1111111111111
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{3.1111111111111}}{2*-1}=\frac{0-\sqrt{3.1111111111111}}{-2}
=-\frac{\sqrt{}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{3.1111111111111}}{2*-1}=\frac{0+\sqrt{3.1111111111111}}{-2}
=\frac{\sqrt{}}{-2}
$