F(4)=a^2-4a-5

Solution for F(4)=a^2-4a-5 equation:

(4)=F^2-4F-5

We move all terms to the left:

(4)-(F^2-4F-5)=0

We get rid of parentheses

-F^2+4F+5+4=0

We add all the numbers together, and all the variables

-1F^2+4F+9=0

a = -1; b = 4; c = +9;
Δ = b2-4ac
Δ = 42-4·(-1)·9
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{13}}{2*-1}=\frac{-4-2\sqrt{13}}{-2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{13}}{2*-1}=\frac{-4+2\sqrt{13}}{-2}
$