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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -2; b = 6; c = -1;
Δ = b2-4ac
Δ = 62-4·(-2)·(-1)
Δ = 28
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:

$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{7}}{2*-2}=\frac{-6-2\sqrt{7}}{-4}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{7}}{2*-2}=\frac{-6+2\sqrt{7}}{-4}
$