(3x+4)(3x-4)=(6x-7)(3x+2)

Solution for (3x+4)(3x-4)=(6x-7)(3x+2) equation:

(3x+4)(3x-4)=(6x-7)(3x+2)

We move all terms to the left:

(3x+4)(3x-4)-((6x-7)(3x+2))=0

We use the square of the difference formula

9x^2-((6x-7)(3x+2))-16=0

We multiply parentheses ..

9x^2-((+18x^2+12x-21x-14))-16=0


We calculate terms in parentheses: -((+18x^2+12x-21x-14)), so:

(+18x^2+12x-21x-14)

We get rid of parentheses

18x^2+12x-21x-14

We add all the numbers together, and all the variables

18x^2-9x-14

Back to the equation:

-(18x^2-9x-14)


We get rid of parentheses

9x^2-18x^2+9x+14-16=0

We add all the numbers together, and all the variables

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

a = -9; b = 9; c = -2;
Δ = b2-4ac
Δ = 92-4·(-9)·(-2)
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3}{2*-9}=\frac{-12}{-18}
=2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3}{2*-9}=\frac{-6}{-18}
=1/3
$