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

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

(3)=5F^2+4F

We move all terms to the left:

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

We get rid of parentheses

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

a = -5; b = -4; c = +3;
Δ = b2-4ac
Δ = -42-4·(-5)·3
Δ = 76
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:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{19}}{2*-5}=\frac{4-2\sqrt{19}}{-10}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{19}}{2*-5}=\frac{4+2\sqrt{19}}{-10}
$