F(x)=(3x+2)(4x+8)

Solution for F(x)=(3x+2)(4x+8) equation:

(F)=(3F+2)(4F+8)

We move all terms to the left:

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

We multiply parentheses ..

-((+12F^2+24F+8F+16))+F=0


We calculate terms in parentheses: -((+12F^2+24F+8F+16)), so:

(+12F^2+24F+8F+16)

We get rid of parentheses

12F^2+24F+8F+16

We add all the numbers together, and all the variables

12F^2+32F+16

Back to the equation:

-(12F^2+32F+16)


We add all the numbers together, and all the variables

F-(12F^2+32F+16)=0

We get rid of parentheses

-12F^2+F-32F-16=0

We add all the numbers together, and all the variables

-12F^2-31F-16=0

a = -12; b = -31; c = -16;
Δ = b2-4ac
Δ = -312-4·(-12)·(-16)
Δ = 193
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{-(-31)-\sqrt{193}}{2*-12}=\frac{31-\sqrt{193}}{-24}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+\sqrt{193}}{2*-12}=\frac{31+\sqrt{193}}{-24}
$