(5x)^2+(3x)^2+(2x)^2=5472

Solution for (5x)^2+(3x)^2+(2x)^2=5472 equation:

(5x)^2+(3x)^2+(2x)^2=5472

We move all terms to the left:

(5x)^2+(3x)^2+(2x)^2-(5472)=0

We add all the numbers together, and all the variables

10x^2-5472=0

a = 10; b = 0; c = -5472;
Δ = b2-4ac
Δ = 02-4·10·(-5472)
Δ = 218880
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:

$\sqrt{\Delta}=\sqrt{218880}=\sqrt{2304*95}=\sqrt{2304}*\sqrt{95}=48\sqrt{95}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48\sqrt{95}}{2*10}=\frac{0-48\sqrt{95}}{20}
=-\frac{48\sqrt{95}}{20}
=-\frac{12\sqrt{95}}{5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48\sqrt{95}}{2*10}=\frac{0+48\sqrt{95}}{20}
=\frac{48\sqrt{95}}{20}
=\frac{12\sqrt{95}}{5}
$