F(x)=9x^2+2x+3x

Solution for F(x)=9x^2+2x+3x equation:

(F)=9F^2+2F+3F

We move all terms to the left:

(F)-(9F^2+2F+3F)=0

We get rid of parentheses

-9F^2+F-2F-3F=0

We add all the numbers together, and all the variables

-9F^2-4F=0

a = -9; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·(-9)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*-9}=\frac{0}{-18}
=0
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*-9}=\frac{8}{-18}
=-4/9
$