(3x+1)(x-3)/x+4=0

Solution for (3x+1)(x-3)/x+4=0 equation:

(3x+1)(x-3)/x+4=0


Domain of the equation: x!=0

x∈R


We multiply parentheses ..

(+3x^2-9x+x-3)/x+4=0

We multiply all the terms by the denominator


(+3x^2-9x+x-3)+4*x=0

We add all the numbers together, and all the variables

(+3x^2-9x+x-3)+4x=0

We get rid of parentheses

3x^2-9x+x+4x-3=0

We add all the numbers together, and all the variables

3x^2-4x-3=0

a = 3; b = -4; c = -3;
Δ = b2-4ac
Δ = -42-4·3·(-3)
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{13}}{2*3}=\frac{4-2\sqrt{13}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{13}}{2*3}=\frac{4+2\sqrt{13}}{6}
$