9x(x+1)=6(x+3)

Solution for 9x(x+1)=6(x+3) equation:

9x(x+1)=6(x+3)

We move all terms to the left:

9x(x+1)-(6(x+3))=0

We multiply parentheses

9x^2+9x-(6(x+3))=0


We calculate terms in parentheses: -(6(x+3)), so:

6(x+3)

We multiply parentheses

6x+18

Back to the equation:

-(6x+18)


We get rid of parentheses

9x^2+9x-6x-18=0

We add all the numbers together, and all the variables

9x^2+3x-18=0

a = 9; b = 3; c = -18;
Δ = b2-4ac
Δ = 32-4·9·(-18)
Δ = 657
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:

$\sqrt{\Delta}=\sqrt{657}=\sqrt{9*73}=\sqrt{9}*\sqrt{73}=3\sqrt{73}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{73}}{2*9}=\frac{-3-3\sqrt{73}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{73}}{2*9}=\frac{-3+3\sqrt{73}}{18}
$