(6x^2+19x+7)/(3x+5)=x

Solution for (6x^2+19x+7)/(3x+5)=x equation:

(6x^2+19x+7)/(3x+5)=x

We move all terms to the left:

(6x^2+19x+7)/(3x+5)-(x)=0


Domain of the equation: (3x+5)!=0

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

3x!=-5

x!=-5/3

x!=-1+2/3

x∈R


We add all the numbers together, and all the variables

-1x+(6x^2+19x+7)/(3x+5)=0

We multiply all the terms by the denominator


-1x*(3x+5)+(6x^2+19x+7)=0

We multiply parentheses

-3x^2-5x+(6x^2+19x+7)=0

We get rid of parentheses

-3x^2+6x^2-5x+19x+7=0

We add all the numbers together, and all the variables

3x^2+14x+7=0

a = 3; b = 14; c = +7;
Δ = b2-4ac
Δ = 142-4·3·7
Δ = 112
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-4\sqrt{7}}{2*3}=\frac{-14-4\sqrt{7}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+4\sqrt{7}}{2*3}=\frac{-14+4\sqrt{7}}{6}
$