(3x-2)(x-8)=(1-2x)(x+6)

Solution for (3x-2)(x-8)=(1-2x)(x+6) equation:

(3x-2)(x-8)=(1-2x)(x+6)

We move all terms to the left:

(3x-2)(x-8)-((1-2x)(x+6))=0

We add all the numbers together, and all the variables

(3x-2)(x-8)-((-2x+1)(x+6))=0

We multiply parentheses ..

(+3x^2-24x-2x+16)-((-2x+1)(x+6))=0


We calculate terms in parentheses: -((-2x+1)(x+6)), so:

(-2x+1)(x+6)

We multiply parentheses ..

(-2x^2-12x+x+6)

We get rid of parentheses

-2x^2-12x+x+6

We add all the numbers together, and all the variables

-2x^2-11x+6

Back to the equation:

-(-2x^2-11x+6)


We get rid of parentheses

3x^2+2x^2-24x-2x+11x+16-6=0

We add all the numbers together, and all the variables

5x^2-15x+10=0

a = 5; b = -15; c = +10;
Δ = b2-4ac
Δ = -152-4·5·10
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-5}{2*5}=\frac{10}{10}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+5}{2*5}=\frac{20}{10}
=2
$