F(x)=3x2+18x+14

Solution for F(x)=3x2+18x+14 equation:

(F)=3F^2+18F+14

We move all terms to the left:

(F)-(3F^2+18F+14)=0

We get rid of parentheses

-3F^2+F-18F-14=0

We add all the numbers together, and all the variables

-3F^2-17F-14=0

a = -3; b = -17; c = -14;
Δ = b2-4ac
Δ = -172-4·(-3)·(-14)
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{-(-17)-11}{2*-3}=\frac{6}{-6}
=-1
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+11}{2*-3}=\frac{28}{-6}
=-4+2/3
$