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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

12x-(5(2x+6)2x)+12=0


We calculate terms in parentheses: -(5(2x+6)2x), so:

5(2x+6)2x

We multiply parentheses

20x^2+60x

Back to the equation:

-(20x^2+60x)


We get rid of parentheses

-20x^2+12x-60x+12=0

We add all the numbers together, and all the variables

-20x^2-48x+12=0

a = -20; b = -48; c = +12;
Δ = b2-4ac
Δ = -482-4·(-20)·12
Δ = 3264
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:

$\sqrt{\Delta}=\sqrt{3264}=\sqrt{64*51}=\sqrt{64}*\sqrt{51}=8\sqrt{51}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-8\sqrt{51}}{2*-20}=\frac{48-8\sqrt{51}}{-40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+8\sqrt{51}}{2*-20}=\frac{48+8\sqrt{51}}{-40}
$