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

Solution for (4x+1)(2-x)=(2-x)(5+3x) equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-4x^2+8x-1x+2)-((-1x+2)(3x+5))=0


We calculate terms in parentheses: -((-1x+2)(3x+5)), so:

(-1x+2)(3x+5)

We multiply parentheses ..

(-3x^2-5x+6x+10)

We get rid of parentheses

-3x^2-5x+6x+10

We add all the numbers together, and all the variables

-3x^2+x+10

Back to the equation:

-(-3x^2+x+10)


We get rid of parentheses

-4x^2+3x^2+8x-1x-x+2-10=0

We add all the numbers together, and all the variables

-1x^2+6x-8=0

a = -1; b = 6; c = -8;
Δ = b2-4ac
Δ = 62-4·(-1)·(-8)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2}{2*-1}=\frac{-8}{-2}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2}{2*-1}=\frac{-4}{-2}
=+2
$