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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-4x^2+8x-6x+12)-((3x+6))=0


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

(3x+6)

We get rid of parentheses

3x+6

Back to the equation:

-(3x+6)


We get rid of parentheses

-4x^2+8x-6x-3x+12-6=0

We add all the numbers together, and all the variables

-4x^2-1x+6=0

a = -4; b = -1; c = +6;
Δ = b2-4ac
Δ = -12-4·(-4)·6
Δ = 97
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{97}}{2*-4}=\frac{1-\sqrt{97}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{97}}{2*-4}=\frac{1+\sqrt{97}}{-8}
$