F(12)=2/3x+9x

Solution for F(12)=2/3x+9x equation:

(12)=2/3F+9F

We move all terms to the left:

(12)-(2/3F+9F)=0


Domain of the equation: 3F+9F)!=0

F∈R


We add all the numbers together, and all the variables

-(+9F+2/3F)+12=0

We get rid of parentheses

-9F-2/3F+12=0

We multiply all the terms by the denominator


-9F*3F+12*3F-2=0

Wy multiply elements

-27F^2+36F-2=0

a = -27; b = 36; c = -2;
Δ = b2-4ac
Δ = 362-4·(-27)·(-2)
Δ = 1080
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:

$\sqrt{\Delta}=\sqrt{1080}=\sqrt{36*30}=\sqrt{36}*\sqrt{30}=6\sqrt{30}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-6\sqrt{30}}{2*-27}=\frac{-36-6\sqrt{30}}{-54}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+6\sqrt{30}}{2*-27}=\frac{-36+6\sqrt{30}}{-54}
$