(3x-4)(x-2)=2x(x-4)+11

Solution for (3x-4)(x-2)=2x(x-4)+11 equation:

(3x-4)(x-2)=2x(x-4)+11

We move all terms to the left:

(3x-4)(x-2)-(2x(x-4)+11)=0

We multiply parentheses ..

(+3x^2-6x-4x+8)-(2x(x-4)+11)=0


We calculate terms in parentheses: -(2x(x-4)+11), so:

2x(x-4)+11

We multiply parentheses

2x^2-8x+11

Back to the equation:

-(2x^2-8x+11)


We get rid of parentheses

3x^2-2x^2-6x-4x+8x+8-11=0

We add all the numbers together, and all the variables

x^2-2x-3=0

a = 1; b = -2; c = -3;
Δ = b2-4ac
Δ = -22-4·1·(-3)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-4}{2*1}=\frac{-2}{2}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+4}{2*1}=\frac{6}{2}
=3
$