9^2+x^2=(3x)^2

Solution for 9^2+x^2=(3x)^2 equation:

9^2+x^2=(3x)^2

We move all terms to the left:

9^2+x^2-((3x)^2)=0

determiningTheFunctionDomain
x^2-3x^2+9^2=0

We add all the numbers together, and all the variables

-2x^2+81=0

a = -2; b = 0; c = +81;
Δ = b2-4ac
Δ = 02-4·(-2)·81
Δ = 648
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:

$\sqrt{\Delta}=\sqrt{648}=\sqrt{324*2}=\sqrt{324}*\sqrt{2}=18\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18\sqrt{2}}{2*-2}=\frac{0-18\sqrt{2}}{-4}
=-\frac{18\sqrt{2}}{-4}
=-\frac{9\sqrt{2}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18\sqrt{2}}{2*-2}=\frac{0+18\sqrt{2}}{-4}
=\frac{18\sqrt{2}}{-4}
=\frac{9\sqrt{2}}{-2}
$