(4x+3)/(x^2+3x+2)=13/12

Solution for (4x+3)/(x^2+3x+2)=13/12 equation:

(4x+3)/(x^2+3x+2)=13/12

We move all terms to the left:

(4x+3)/(x^2+3x+2)-(13/12)=0


Domain of the equation: (x^2+3x+2)!=0

We move all terms containing x to the left, all other terms to the right

x^2+3x!=-2

x∈R


We add all the numbers together, and all the variables

(4x+3)/(x^2+3x+2)-(+13/12)=0

We get rid of parentheses

(4x+3)/(x^2+3x+2)-13/12=0

We calculate fractions


(-13x^2-39x-26)/(12x^2+36x+24)+(48x+36)/(12x^2+36x+24)=0

We multiply all the terms by the denominator


(-13x^2-39x-26)+(48x+36)=0

We get rid of parentheses

-13x^2-39x+48x-26+36=0

We add all the numbers together, and all the variables

-13x^2+9x+10=0

a = -13; b = 9; c = +10;
Δ = b2-4ac
Δ = 92-4·(-13)·10
Δ = 601
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-\sqrt{601}}{2*-13}=\frac{-9-\sqrt{601}}{-26}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{601}}{2*-13}=\frac{-9+\sqrt{601}}{-26}
$