x=5/3x+7-4(x+1

Solution for x=5/3x+7-4(x+1 equation:

x=5/3x+7-4(x+1

We move all terms to the left:

x-(5/3x+7-4(x+1)=0


Domain of the equation: 3x+7-4(x+1)!=0

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

3x-4(x+1)!=-7

x∈R


We multiply all the terms by the denominator


x*3x-4(x+1)+7-(5=0

We add all the numbers together, and all the variables

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

We multiply parentheses

x*3x-4x-4=0

Wy multiply elements

3x^2-4x-4=0

a = 3; b = -4; c = -4;
Δ = b2-4ac
Δ = -42-4·3·(-4)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8}{2*3}=\frac{-4}{6}
=-2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*3}=\frac{12}{6}
=2
$