F(-1)=x^2+3x-5

Solution for F(-1)=x^2+3x-5 equation:

(-1)=F^2+3F-5

We move all terms to the left:

(-1)-(F^2+3F-5)=0

We add all the numbers together, and all the variables

-(F^2+3F-5)-1=0

We get rid of parentheses

-F^2-3F+5-1=0

We add all the numbers together, and all the variables

-1F^2-3F+4=0

a = -1; b = -3; c = +4;
Δ = b2-4ac
Δ = -32-4·(-1)·4
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-5}{2*-1}=\frac{-2}{-2}
=1
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+5}{2*-1}=\frac{8}{-2}
=-4
$