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

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

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

We move all terms to the left:

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

We multiply parentheses ..

-((+6F^2-3F+2F-1))+F=0


We calculate terms in parentheses: -((+6F^2-3F+2F-1)), so:

(+6F^2-3F+2F-1)

We get rid of parentheses

6F^2-3F+2F-1

We add all the numbers together, and all the variables

6F^2-1F-1

Back to the equation:

-(6F^2-1F-1)


We add all the numbers together, and all the variables

F-(6F^2-1F-1)=0

We get rid of parentheses

-6F^2+F+1F+1=0

We add all the numbers together, and all the variables

-6F^2+2F+1=0

a = -6; b = 2; c = +1;
Δ = b2-4ac
Δ = 22-4·(-6)·1
Δ = 28
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{7}}{2*-6}=\frac{-2-2\sqrt{7}}{-12}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{7}}{2*-6}=\frac{-2+2\sqrt{7}}{-12}
$