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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

9x^2-40=0

a = 9; b = 0; c = -40;
Δ = b2-4ac
Δ = 02-4·9·(-40)
Δ = 1440
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:

$\sqrt{\Delta}=\sqrt{1440}=\sqrt{144*10}=\sqrt{144}*\sqrt{10}=12\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{10}}{2*9}=\frac{0-12\sqrt{10}}{18}
=-\frac{12\sqrt{10}}{18}
=-\frac{2\sqrt{10}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{10}}{2*9}=\frac{0+12\sqrt{10}}{18}
=\frac{12\sqrt{10}}{18}
=\frac{2\sqrt{10}}{3}
$