2/3f+2=f+1

Solution for 2/3f+2=f+1 equation:

2/3f+2=f+1

We move all terms to the left:

2/3f+2-(f+1)=0


Domain of the equation: 3f!=0

f!=0/3

f!=0

f∈R


We get rid of parentheses

2/3f-f-1+2=0

We multiply all the terms by the denominator


-f*3f-1*3f+2*3f+2=0

Wy multiply elements

-3f^2-3f+6f+2=0

We add all the numbers together, and all the variables

-3f^2+3f+2=0

a = -3; b = 3; c = +2;
Δ = b2-4ac
Δ = 32-4·(-3)·2
Δ = 33
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{-(3)-\sqrt{33}}{2*-3}=\frac{-3-\sqrt{33}}{-6}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{33}}{2*-3}=\frac{-3+\sqrt{33}}{-6}
$