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

Solution for 3x(3x-1)=4(3x+1) equation:

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

We move all terms to the left:

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

We multiply parentheses

9x^2-3x-(4(3x+1))=0


We calculate terms in parentheses: -(4(3x+1)), so:

4(3x+1)

We multiply parentheses

12x+4

Back to the equation:

-(12x+4)


We get rid of parentheses

9x^2-3x-12x-4=0

We add all the numbers together, and all the variables

9x^2-15x-4=0

a = 9; b = -15; c = -4;
Δ = b2-4ac
Δ = -152-4·9·(-4)
Δ = 369
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:

$\sqrt{\Delta}=\sqrt{369}=\sqrt{9*41}=\sqrt{9}*\sqrt{41}=3\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-3\sqrt{41}}{2*9}=\frac{15-3\sqrt{41}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+3\sqrt{41}}{2*9}=\frac{15+3\sqrt{41}}{18}
$