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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply parentheses ..

(-3x^2+6x-2x+4)/3-x-1=0

We multiply all the terms by the denominator


(-3x^2+6x-2x+4)-x*3-1*3=0

We add all the numbers together, and all the variables

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

Wy multiply elements

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-3x^2+x+1=0

a = -3; b = 1; c = +1;
Δ = b2-4ac
Δ = 12-4·(-3)·1
Δ = 13
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{13}}{2*-3}=\frac{-1-\sqrt{13}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{13}}{2*-3}=\frac{-1+\sqrt{13}}{-6}
$