F(x)=3x^2+20x

Solution for F(x)=3x^2+20x equation:

(F)=3F^2+20F

We move all terms to the left:

(F)-(3F^2+20F)=0

We get rid of parentheses

-3F^2+F-20F=0

We add all the numbers together, and all the variables

-3F^2-19F=0

a = -3; b = -19; c = 0;
Δ = b2-4ac
Δ = -192-4·(-3)·0
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-19}{2*-3}=\frac{0}{-6}
=0
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+19}{2*-3}=\frac{38}{-6}
=-6+1/3
$