(X-3)(x+4)-2(3x-2)=(x-4)

Solution for (X-3)(x+4)-2(3x-2)=(x-4) equation:

(X-3)(X+4)-2(3X-2)=(X-4)

We move all terms to the left:

(X-3)(X+4)-2(3X-2)-((X-4))=0

We multiply parentheses

(X-3)(X+4)-6X-((X-4))+4=0

We multiply parentheses ..

(+X^2+4X-3X-12)-6X-((X-4))+4=0


We calculate terms in parentheses: -((X-4)), so:

(X-4)

We get rid of parentheses

X-4

Back to the equation:

-(X-4)


We get rid of parentheses

X^2+4X-3X-6X-X-12+4+4=0

We add all the numbers together, and all the variables

X^2-6X-4=0

a = 1; b = -6; c = -4;
Δ = b2-4ac
Δ = -62-4·1·(-4)
Δ = 52
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{13}}{2*1}=\frac{6-2\sqrt{13}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{13}}{2*1}=\frac{6+2\sqrt{13}}{2}
$