(5x^2-x+5)+(x^2+x-5)=1

Solution for (5x^2-x+5)+(x^2+x-5)=1 equation:

(5x^2-x+5)+(x^2+x-5)=1

We move all terms to the left:

(5x^2-x+5)+(x^2+x-5)-(1)=0

We get rid of parentheses

5x^2+x^2-x+x+5-5-1=0

We add all the numbers together, and all the variables

6x^2-1=0

a = 6; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·6·(-1)
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{6}}{2*6}=\frac{0-2\sqrt{6}}{12}
=-\frac{2\sqrt{6}}{12}
=-\frac{\sqrt{6}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{6}}{2*6}=\frac{0+2\sqrt{6}}{12}
=\frac{2\sqrt{6}}{12}
=\frac{\sqrt{6}}{6}
$