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

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

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

We move all terms to the left:

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

We multiply parentheses ..

-((+6F^2+8F-12F-16))+F=0


We calculate terms in parentheses: -((+6F^2+8F-12F-16)), so:

(+6F^2+8F-12F-16)

We get rid of parentheses

6F^2+8F-12F-16

We add all the numbers together, and all the variables

6F^2-4F-16

Back to the equation:

-(6F^2-4F-16)


We add all the numbers together, and all the variables

F-(6F^2-4F-16)=0

We get rid of parentheses

-6F^2+F+4F+16=0

We add all the numbers together, and all the variables

-6F^2+5F+16=0

a = -6; b = 5; c = +16;
Δ = b2-4ac
Δ = 52-4·(-6)·16
Δ = 409
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{409}}{2*-6}=\frac{-5-\sqrt{409}}{-12}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{409}}{2*-6}=\frac{-5+\sqrt{409}}{-12}
$