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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


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

2(+F^2+3F+F+3)

We multiply parentheses

2F^2+6F+2F+6

We add all the numbers together, and all the variables

2F^2+8F+6

Back to the equation:

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


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-2F^2-7F-6=0

a = -2; b = -7; c = -6;
Δ = b2-4ac
Δ = -72-4·(-2)·(-6)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-1}{2*-2}=\frac{6}{-4}
=-1+1/2
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+1}{2*-2}=\frac{8}{-4}
=-2
$