3x(x-7)=(x+5)(x-3)

Solution for 3x(x-7)=(x+5)(x-3) equation:

3x(x-7)=(x+5)(x-3)

We move all terms to the left:

3x(x-7)-((x+5)(x-3))=0

We multiply parentheses

3x^2-21x-((x+5)(x-3))=0

We multiply parentheses ..

3x^2-((+x^2-3x+5x-15))-21x=0


We calculate terms in parentheses: -((+x^2-3x+5x-15)), so:

(+x^2-3x+5x-15)

We get rid of parentheses

x^2-3x+5x-15

We add all the numbers together, and all the variables

x^2+2x-15

Back to the equation:

-(x^2+2x-15)


We add all the numbers together, and all the variables

3x^2-21x-(x^2+2x-15)=0

We get rid of parentheses

3x^2-x^2-21x-2x+15=0

We add all the numbers together, and all the variables

2x^2-23x+15=0

a = 2; b = -23; c = +15;
Δ = b2-4ac
Δ = -232-4·2·15
Δ = 409
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{-(-23)-\sqrt{409}}{2*2}=\frac{23-\sqrt{409}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{409}}{2*2}=\frac{23+\sqrt{409}}{4}
$