(7x-3)(2x+1)=(7x-3)

Solution for (7x-3)(2x+1)=(7x-3) equation:

(7x-3)(2x+1)=(7x-3)

We move all terms to the left:

(7x-3)(2x+1)-((7x-3))=0

We multiply parentheses ..

(+14x^2+7x-6x-3)-((7x-3))=0


We calculate terms in parentheses: -((7x-3)), so:

(7x-3)

We get rid of parentheses

7x-3

Back to the equation:

-(7x-3)


We get rid of parentheses

14x^2+7x-6x-7x-3+3=0

We add all the numbers together, and all the variables

14x^2-6x=0

a = 14; b = -6; c = 0;
Δ = b2-4ac
Δ = -62-4·14·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6}{2*14}=\frac{0}{28}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6}{2*14}=\frac{12}{28}
=3/7
$