F(10)=2x^2+x+4

Solution for F(10)=2x^2+x+4 equation:

(10)=2F^2+F+4

We move all terms to the left:

(10)-(2F^2+F+4)=0

We get rid of parentheses

-2F^2-F-4+10=0

We add all the numbers together, and all the variables

-2F^2-1F+6=0

a = -2; b = -1; c = +6;
Δ = b2-4ac
Δ = -12-4·(-2)·6
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-7}{2*-2}=\frac{-6}{-4}
=1+1/2
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+7}{2*-2}=\frac{8}{-4}
=-2
$