F(x)=(x+3)(x+12)-(x+6)*2

Solution for F(x)=(x+3)(x+12)-(x+6)*2 equation:

(F)=(F+3)(F+12)-(F+6)*2

We move all terms to the left:

(F)-((F+3)(F+12)-(F+6)*2)=0

We multiply parentheses ..

-((+F^2+12F+3F+36)-(F+6)*2)+F=0


We calculate terms in parentheses: -((+F^2+12F+3F+36)-(F+6)*2), so:

(+F^2+12F+3F+36)-(F+6)*2

We multiply parentheses

(+F^2+12F+3F+36)-2F-12

We get rid of parentheses

F^2+12F+3F-2F+36-12

We add all the numbers together, and all the variables

F^2+13F+24

Back to the equation:

-(F^2+13F+24)


We add all the numbers together, and all the variables

F-(F^2+13F+24)=0

We get rid of parentheses

-F^2+F-13F-24=0

We add all the numbers together, and all the variables

-1F^2-12F-24=0

a = -1; b = -12; c = -24;
Δ = b2-4ac
Δ = -122-4·(-1)·(-24)
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{3}}{2*-1}=\frac{12-4\sqrt{3}}{-2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{3}}{2*-1}=\frac{12+4\sqrt{3}}{-2}
$