F(x)=3x^2-24x+42

Solution for F(x)=3x^2-24x+42 equation:

(F)=3F^2-24F+42

We move all terms to the left:

(F)-(3F^2-24F+42)=0

We get rid of parentheses

-3F^2+F+24F-42=0

We add all the numbers together, and all the variables

-3F^2+25F-42=0

a = -3; b = 25; c = -42;
Δ = b2-4ac
Δ = 252-4·(-3)·(-42)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-11}{2*-3}=\frac{-36}{-6}
=+6
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+11}{2*-3}=\frac{-14}{-6}
=2+1/3
$