(3x)^2+(4x)^2+(5x)^2=1250

Solution for (3x)^2+(4x)^2+(5x)^2=1250 equation:

(3x)^2+(4x)^2+(5x)^2=1250

We move all terms to the left:

(3x)^2+(4x)^2+(5x)^2-(1250)=0

We add all the numbers together, and all the variables

12x^2-1250=0

a = 12; b = 0; c = -1250;
Δ = b2-4ac
Δ = 02-4·12·(-1250)
Δ = 60000
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:

$\sqrt{\Delta}=\sqrt{60000}=\sqrt{10000*6}=\sqrt{10000}*\sqrt{6}=100\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-100\sqrt{6}}{2*12}=\frac{0-100\sqrt{6}}{24}
=-\frac{100\sqrt{6}}{24}
=-\frac{25\sqrt{6}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+100\sqrt{6}}{2*12}=\frac{0+100\sqrt{6}}{24}
=\frac{100\sqrt{6}}{24}
=\frac{25\sqrt{6}}{6}
$