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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -2; b = 6; c = -2;
Δ = b2-4ac
Δ = 62-4·(-2)·(-2)
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{5}}{2*-2}=\frac{-6-2\sqrt{5}}{-4}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{5}}{2*-2}=\frac{-6+2\sqrt{5}}{-4}
$