(3x+1)(5x+2)=(6x+2)(1-x)

Solution for (3x+1)(5x+2)=(6x+2)(1-x) equation:

(3x+1)(5x+2)=(6x+2)(1-x)

We move all terms to the left:

(3x+1)(5x+2)-((6x+2)(1-x))=0

We add all the numbers together, and all the variables

(3x+1)(5x+2)-((6x+2)(-1x+1))=0

We multiply parentheses ..

(+15x^2+6x+5x+2)-((6x+2)(-1x+1))=0


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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-6x^2+4x+2

Back to the equation:

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


We get rid of parentheses

15x^2+6x^2+6x+5x-4x+2-2=0

We add all the numbers together, and all the variables

21x^2+7x=0

a = 21; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·21·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-7}{2*21}=\frac{-14}{42}
=-1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*21}=\frac{0}{42}
=0
$