F(2)=5x^2-4x

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

(2)=5F^2-4F

We move all terms to the left:

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

We get rid of parentheses

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

a = -5; b = 4; c = +2;
Δ = b2-4ac
Δ = 42-4·(-5)·2
Δ = 56
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:

$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{14}}{2*-5}=\frac{-4-2\sqrt{14}}{-10}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{14}}{2*-5}=\frac{-4+2\sqrt{14}}{-10}
$