(6x^2+13x+5)/(2x+1)=x

Solution for (6x^2+13x+5)/(2x+1)=x equation:

(6x^2+13x+5)/(2x+1)=x

We move all terms to the left:

(6x^2+13x+5)/(2x+1)-(x)=0


Domain of the equation: (2x+1)!=0

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

2x!=-1

x!=-1/2

x!=-1/2

x∈R


We add all the numbers together, and all the variables

-1x+(6x^2+13x+5)/(2x+1)=0

We multiply all the terms by the denominator


-1x*(2x+1)+(6x^2+13x+5)=0

We multiply parentheses

-2x^2-x+(6x^2+13x+5)=0

We get rid of parentheses

-2x^2+6x^2-x+13x+5=0

We add all the numbers together, and all the variables

4x^2+12x+5=0

a = 4; b = 12; c = +5;
Δ = b2-4ac
Δ = 122-4·4·5
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8}{2*4}=\frac{-20}{8}
=-2+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8}{2*4}=\frac{-4}{8}
=-1/2
$