F(3)=x2-5

Solution for F(3)=x2-5 equation:

(3)=F2-5

We move all terms to the left:

(3)-(F2-5)=0

We add all the numbers together, and all the variables

-(+F^2-5)+3=0

We get rid of parentheses

-F^2+5+3=0

We add all the numbers together, and all the variables

-1F^2+8=0

a = -1; b = 0; c = +8;
Δ = b2-4ac
Δ = 02-4·(-1)·8
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{2}}{2*-1}=\frac{0-4\sqrt{2}}{-2}
=-\frac{4\sqrt{2}}{-2}
=-\frac{2\sqrt{2}}{-1}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{2}}{2*-1}=\frac{0+4\sqrt{2}}{-2}
=\frac{4\sqrt{2}}{-2}
=\frac{2\sqrt{2}}{-1}
$