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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -2; b = -5; c = -3;
Δ = b2-4ac
Δ = -52-4·(-2)·(-3)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-1}{2*-2}=\frac{4}{-4}
=-1
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+1}{2*-2}=\frac{6}{-4}
=-1+1/2
$