F(x)=3(x-2)(x+4)

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

(F)=3(F-2)(F+4)

We move all terms to the left:

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

We multiply parentheses ..

-(3(+F^2+4F-2F-8))+F=0


We calculate terms in parentheses: -(3(+F^2+4F-2F-8)), so:

3(+F^2+4F-2F-8)

We multiply parentheses

3F^2+12F-6F-24

We add all the numbers together, and all the variables

3F^2+6F-24

Back to the equation:

-(3F^2+6F-24)


We add all the numbers together, and all the variables

F-(3F^2+6F-24)=0

We get rid of parentheses

-3F^2+F-6F+24=0

We add all the numbers together, and all the variables

-3F^2-5F+24=0

a = -3; b = -5; c = +24;
Δ = b2-4ac
Δ = -52-4·(-3)·24
Δ = 313
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{313}}{2*-3}=\frac{5-\sqrt{313}}{-6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{313}}{2*-3}=\frac{5+\sqrt{313}}{-6}
$