F(x)=(3x+1)(2x-8)

Solution for F(x)=(3x+1)(2x-8) equation:

(F)=(3F+1)(2F-8)

We move all terms to the left:

(F)-((3F+1)(2F-8))=0

We multiply parentheses ..

-((+6F^2-24F+2F-8))+F=0


We calculate terms in parentheses: -((+6F^2-24F+2F-8)), so:

(+6F^2-24F+2F-8)

We get rid of parentheses

6F^2-24F+2F-8

We add all the numbers together, and all the variables

6F^2-22F-8

Back to the equation:

-(6F^2-22F-8)


We add all the numbers together, and all the variables

F-(6F^2-22F-8)=0

We get rid of parentheses

-6F^2+F+22F+8=0

We add all the numbers together, and all the variables

-6F^2+23F+8=0

a = -6; b = 23; c = +8;
Δ = b2-4ac
Δ = 232-4·(-6)·8
Δ = 721
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{-(23)-\sqrt{721}}{2*-6}=\frac{-23-\sqrt{721}}{-12}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{721}}{2*-6}=\frac{-23+\sqrt{721}}{-12}
$