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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -2; b = 4; c = -1;
Δ = b2-4ac
Δ = 42-4·(-2)·(-1)
Δ = 8
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:

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