F(4)=2x^2+5x+3

Solution for F(4)=2x^2+5x+3 equation:

(4)=2F^2+5F+3

We move all terms to the left:

(4)-(2F^2+5F+3)=0

We get rid of parentheses

-2F^2-5F-3+4=0

We add all the numbers together, and all the variables

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

a = -2; b = -5; c = +1;
Δ = b2-4ac
Δ = -52-4·(-2)·1
Δ = 33
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{33}}{2*-2}=\frac{5-\sqrt{33}}{-4}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{33}}{2*-2}=\frac{5+\sqrt{33}}{-4}
$