(3-2x)(x+4)=(x+5)(2-x)-2x

Solution for (3-2x)(x+4)=(x+5)(2-x)-2x equation:

(3-2x)(x+4)=(x+5)(2-x)-2x

We move all terms to the left:

(3-2x)(x+4)-((x+5)(2-x)-2x)=0

We add all the numbers together, and all the variables

(-2x+3)(x+4)-((x+5)(-1x+2)-2x)=0

We multiply parentheses ..

(-2x^2-8x+3x+12)-((x+5)(-1x+2)-2x)=0


We calculate terms in parentheses: -((x+5)(-1x+2)-2x), so:

(x+5)(-1x+2)-2x

We add all the numbers together, and all the variables

-2x+(x+5)(-1x+2)

We multiply parentheses ..

(-1x^2+2x-5x+10)-2x

We get rid of parentheses

-1x^2+2x-5x-2x+10

We add all the numbers together, and all the variables

-1x^2-5x+10

Back to the equation:

-(-1x^2-5x+10)


We get rid of parentheses

-2x^2+1x^2-8x+3x+5x+12-10=0

We add all the numbers together, and all the variables

-1x^2+2=0

a = -1; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-1)·2
Δ = 8
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:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{2}}{2*-1}=\frac{0-2\sqrt{2}}{-2}
=-\frac{2\sqrt{2}}{-2}
=-\frac{\sqrt{2}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{2}}{2*-1}=\frac{0+2\sqrt{2}}{-2}
=\frac{2\sqrt{2}}{-2}
=\frac{\sqrt{2}}{-1}
$