(3x)^2+(6x)^2=(40)^2

Solution for (3x)^2+(6x)^2=(40)^2 equation:

(3x)^2+(6x)^2=(40)^2

We move all terms to the left:

(3x)^2+(6x)^2-((40)^2)=0

determiningTheFunctionDomain
3x^2+6x^2-40^2=0

We add all the numbers together, and all the variables

9x^2-1600=0

a = 9; b = 0; c = -1600;
Δ = b2-4ac
Δ = 02-4·9·(-1600)
Δ = 57600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{57600}=240$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-240}{2*9}=\frac{-240}{18}
=-13+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+240}{2*9}=\frac{240}{18}
=13+1/3
$