31958x^2+(362x^2)=123

Solution for 31958x^2+(362x^2)=123 equation:

31958x^2+(362x^2)=123

We move all terms to the left:

31958x^2+(362x^2)-(123)=0

We add all the numbers together, and all the variables

32320x^2-123=0

a = 32320; b = 0; c = -123;
Δ = b2-4ac
Δ = 02-4·32320·(-123)
Δ = 15901440
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:

$\sqrt{\Delta}=\sqrt{15901440}=\sqrt{256*62115}=\sqrt{256}*\sqrt{62115}=16\sqrt{62115}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{62115}}{2*32320}=\frac{0-16\sqrt{62115}}{64640}
=-\frac{16\sqrt{62115}}{64640}
=-\frac{\sqrt{62115}}{4040}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{62115}}{2*32320}=\frac{0+16\sqrt{62115}}{64640}
=\frac{16\sqrt{62115}}{64640}
=\frac{\sqrt{62115}}{4040}
$