4(3x+6)/5=2(2x+5)/3-3x-3

Solution for 4(3x+6)/5=2(2x+5)/3-3x-3 equation:

4(3x+6)/5=2(2x+5)/3-3x-3

We move all terms to the left:

4(3x+6)/5-(2(2x+5)/3-3x-3)=0


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

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

-3x-3)!=-3

x∈R


We calculate fractions


(33x+72)/(-3x)+(-(2(2x+5)*5)/(-3x)=0


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

-(2(2x+5)*5)/(-3x

We multiply all the terms by the denominator


-(2(2x+5)*5)


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

2(2x+5)*5

We multiply parentheses

20x+50

Back to the equation:

-(20x+50)


We get rid of parentheses

-20x-50

Back to the equation:

+(-20x-50)


We get rid of parentheses

(33x+72)/(-3x)-20x-50=0

We multiply all the terms by the denominator


(33x+72)-20x*(-3x)-50*(-3x)=0

We multiply parentheses

60x^2+(33x+72)+150x=0

We get rid of parentheses

60x^2+33x+150x+72=0

We add all the numbers together, and all the variables

60x^2+183x+72=0

a = 60; b = 183; c = +72;
Δ = b2-4ac
Δ = 1832-4·60·72
Δ = 16209
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:

$\sqrt{\Delta}=\sqrt{16209}=\sqrt{9*1801}=\sqrt{9}*\sqrt{1801}=3\sqrt{1801}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(183)-3\sqrt{1801}}{2*60}=\frac{-183-3\sqrt{1801}}{120}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(183)+3\sqrt{1801}}{2*60}=\frac{-183+3\sqrt{1801}}{120}
$