(4x+4)-(7x-8)/4x+5=4/3

Solution for (4x+4)-(7x-8)/4x+5=4/3 equation:

(4x+4)-(7x-8)/4x+5=4/3

We move all terms to the left:

(4x+4)-(7x-8)/4x+5-(4/3)=0


Domain of the equation: 4x!=0

x!=0/4

x!=0

x∈R


We add all the numbers together, and all the variables

(4x+4)-(7x-8)/4x+5-(+4/3)=0

We get rid of parentheses

4x-(7x-8)/4x+4+5-4/3=0

We calculate fractions


4x+(-21x+24)/12x+(-16x)/12x+4+5=0

We add all the numbers together, and all the variables

4x+(-21x+24)/12x+(-16x)/12x+9=0

We multiply all the terms by the denominator


4x*12x+(-21x+24)+(-16x)+9*12x=0

Wy multiply elements

48x^2+(-21x+24)+(-16x)+108x=0

We get rid of parentheses

48x^2-21x-16x+108x+24=0

We add all the numbers together, and all the variables

48x^2+71x+24=0

a = 48; b = 71; c = +24;
Δ = b2-4ac
Δ = 712-4·48·24
Δ = 433
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{-(71)-\sqrt{433}}{2*48}=\frac{-71-\sqrt{433}}{96}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(71)+\sqrt{433}}{2*48}=\frac{-71+\sqrt{433}}{96}
$