(2+3x)(2-3x)=16x

Solution for (2+3x)(2-3x)=16x equation:

(2+3x)(2-3x)=16x

We move all terms to the left:

(2+3x)(2-3x)-(16x)=0

We add all the numbers together, and all the variables

(3x+2)(-3x+2)-16x=0

We add all the numbers together, and all the variables

-16x+(3x+2)(-3x+2)=0

We multiply parentheses ..

(-9x^2+6x-6x+4)-16x=0

We get rid of parentheses

-9x^2+6x-6x-16x+4=0

We add all the numbers together, and all the variables

-9x^2-16x+4=0

a = -9; b = -16; c = +4;
Δ = b2-4ac
Δ = -162-4·(-9)·4
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-20}{2*-9}=\frac{-4}{-18}
=2/9
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+20}{2*-9}=\frac{36}{-18}
=-2
$