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

Solution for (3-4x)(3+4x)=x^2-9(x-1) equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-16x^2-12x+12x+9)-(x^2-9(x-1))=0


We calculate terms in parentheses: -(x^2-9(x-1)), so:

x^2-9(x-1)

We multiply parentheses

x^2-9x+9

Back to the equation:

-(x^2-9x+9)


We get rid of parentheses

-16x^2-x^2-12x+12x+9x+9-9=0

We add all the numbers together, and all the variables

-17x^2+9x=0

a = -17; b = 9; c = 0;
Δ = b2-4ac
Δ = 92-4·(-17)·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-9}{2*-17}=\frac{-18}{-34}
=9/17
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+9}{2*-17}=\frac{0}{-34}
=0
$