(9x+3)(x+1)=(9x+3)(5x-4)

Solution for (9x+3)(x+1)=(9x+3)(5x-4) equation:

(9x+3)(x+1)=(9x+3)(5x-4)

We move all terms to the left:

(9x+3)(x+1)-((9x+3)(5x-4))=0

We multiply parentheses ..

(+9x^2+9x+3x+3)-((9x+3)(5x-4))=0


We calculate terms in parentheses: -((9x+3)(5x-4)), so:

(9x+3)(5x-4)

We multiply parentheses ..

(+45x^2-36x+15x-12)

We get rid of parentheses

45x^2-36x+15x-12

We add all the numbers together, and all the variables

45x^2-21x-12

Back to the equation:

-(45x^2-21x-12)


We get rid of parentheses

9x^2-45x^2+9x+3x+21x+3+12=0

We add all the numbers together, and all the variables

-36x^2+33x+15=0

a = -36; b = 33; c = +15;
Δ = b2-4ac
Δ = 332-4·(-36)·15
Δ = 3249
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3249}=57$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-57}{2*-36}=\frac{-90}{-72}
=1+1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+57}{2*-36}=\frac{24}{-72}
=-1/3
$