((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+4)!=0

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

x!=-4

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))=0


We calculate terms in parentheses: +((+3x^2-9x+x-3)), so:

(+3x^2-9x+x-3)

We get rid of parentheses

3x^2-9x+x-3

We add all the numbers together, and all the variables

3x^2-8x-3

Back to the equation:

+(3x^2-8x-3)


We get rid of parentheses

3x^2-8x-3=0

a = 3; b = -8; c = -3;
Δ = b2-4ac
Δ = -82-4·3·(-3)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-10}{2*3}=\frac{-2}{6}
=-1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+10}{2*3}=\frac{18}{6}
=3
$