f=5X^2-10+X

Solution for f=5X^2-10+X equation:

f=5f^2-10+f

We move all terms to the left:

f-(5f^2-10+f)=0

We get rid of parentheses

-5f^2+f-f+10=0

We add all the numbers together, and all the variables

-5f^2+10=0

a = -5; b = 0; c = +10;
Δ = b2-4ac
Δ = 02-4·(-5)·10
Δ = 200
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{2}}{2*-5}=\frac{0-10\sqrt{2}}{-10}
=-\frac{10\sqrt{2}}{-10}
=-\frac{\sqrt{2}}{-1}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{2}}{2*-5}=\frac{0+10\sqrt{2}}{-10}
=\frac{10\sqrt{2}}{-10}
=\frac{\sqrt{2}}{-1}
$