((5-4x)/3x^2-x-4)=4

Solution for ((5-4x)/3x^2-x-4)=4 equation:

((5-4x)/3x^2-x-4)=4

We move all terms to the left:

((5-4x)/3x^2-x-4)-(4)=0


Domain of the equation: 3x^2-x-4)!=0

x∈R


We add all the numbers together, and all the variables

((-4x+5)/3x^2-x-4)-4=0

We multiply all the terms by the denominator


((-4x+5)-4*3x^2-x-4)=0


We calculate terms in parentheses: +((-4x+5)-4*3x^2-x-4), so:

(-4x+5)-4*3x^2-x-4

We add all the numbers together, and all the variables

-1x+(-4x+5)-4*3x^2-4

Wy multiply elements

-12x^2-1x+(-4x+5)-4

We get rid of parentheses

-12x^2-1x-4x+5-4

We add all the numbers together, and all the variables

-12x^2-5x+1

Back to the equation:

+(-12x^2-5x+1)


We get rid of parentheses

-12x^2-5x+1=0

a = -12; b = -5; c = +1;
Δ = b2-4ac
Δ = -52-4·(-12)·1
Δ = 73
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{-(-5)-\sqrt{73}}{2*-12}=\frac{5-\sqrt{73}}{-24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{73}}{2*-12}=\frac{5+\sqrt{73}}{-24}
$