234x^2+(546x^2)-(352x^2)+(1234x^2)-(8x^2)+(2354x^2)=13

Solution for 234x^2+(546x^2)-(352x^2)+(1234x^2)-(8x^2)+(2354x^2)=13 equation:

234x^2+(546x^2)-(352x^2)+(1234x^2)-(8x^2)+(2354x^2)=13

We move all terms to the left:

234x^2+(546x^2)-(352x^2)+(1234x^2)-(8x^2)+(2354x^2)-(13)=0

We add all the numbers together, and all the variables

4008x^2-13=0

a = 4008; b = 0; c = -13;
Δ = b2-4ac
Δ = 02-4·4008·(-13)
Δ = 208416
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:

$\sqrt{\Delta}=\sqrt{208416}=\sqrt{16*13026}=\sqrt{16}*\sqrt{13026}=4\sqrt{13026}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{13026}}{2*4008}=\frac{0-4\sqrt{13026}}{8016}
=-\frac{4\sqrt{13026}}{8016}
=-\frac{\sqrt{13026}}{2004}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{13026}}{2*4008}=\frac{0+4\sqrt{13026}}{8016}
=\frac{4\sqrt{13026}}{8016}
=\frac{\sqrt{13026}}{2004}
$