x*(-4)=(2x+1)(3-2x)

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((-4x^2+6x-2x+3)), so:

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

We get rid of parentheses

-4x^2+6x-2x+3

We add all the numbers together, and all the variables

-4x^2+4x+3

Back to the equation:

-(-4x^2+4x+3)


We get rid of parentheses

4x^2-4x-4x-3=0

We add all the numbers together, and all the variables

4x^2-8x-3=0

a = 4; b = -8; c = -3;
Δ = b2-4ac
Δ = -82-4·4·(-3)
Δ = 112
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{7}}{2*4}=\frac{8-4\sqrt{7}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{7}}{2*4}=\frac{8+4\sqrt{7}}{8}
$