F(x)=5+X-2X^2

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

(F)=5+F-2F^2

We move all terms to the left:

(F)-(5+F-2F^2)=0

We get rid of parentheses

2F^2-F+F-5=0

We add all the numbers together, and all the variables

2F^2-5=0

a = 2; b = 0; c = -5;
Δ = b2-4ac
Δ = 02-4·2·(-5)
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{10}}{2*2}=\frac{0-2\sqrt{10}}{4}
=-\frac{2\sqrt{10}}{4}
=-\frac{\sqrt{10}}{2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{10}}{2*2}=\frac{0+2\sqrt{10}}{4}
=\frac{2\sqrt{10}}{4}
=\frac{\sqrt{10}}{2}
$