(7x^2+4x-3)=(x-1)(x+3)

Solution for (7x^2+4x-3)=(x-1)(x+3) equation:

(7x^2+4x-3)=(x-1)(x+3)

We move all terms to the left:

(7x^2+4x-3)-((x-1)(x+3))=0

We get rid of parentheses

7x^2+4x-((x-1)(x+3))-3=0

We multiply parentheses ..

7x^2-((+x^2+3x-1x-3))+4x-3=0


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

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

We get rid of parentheses

x^2+3x-1x-3

We add all the numbers together, and all the variables

x^2+2x-3

Back to the equation:

-(x^2+2x-3)


We add all the numbers together, and all the variables

7x^2+4x-(x^2+2x-3)-3=0

We get rid of parentheses

7x^2-x^2+4x-2x+3-3=0

We add all the numbers together, and all the variables

6x^2+2x=0

a = 6; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·6·0
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{-(2)-2}{2*6}=\frac{-4}{12}
=-1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*6}=\frac{0}{12}
=0
$