2x(2x-3)=x^2-(6x-4)

Solution for 2x(2x-3)=x^2-(6x-4) equation:

2x(2x-3)=x^2-(6x-4)

We move all terms to the left:

2x(2x-3)-(x^2-(6x-4))=0

We multiply parentheses

4x^2-6x-(x^2-(6x-4))=0


We calculate terms in parentheses: -(x^2-(6x-4)), so:

x^2-(6x-4)

We get rid of parentheses

x^2-6x+4

Back to the equation:

-(x^2-6x+4)


We get rid of parentheses

4x^2-x^2-6x+6x-4=0

We add all the numbers together, and all the variables

3x^2-4=0

a = 3; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·3·(-4)
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{3}}{2*3}=\frac{0-4\sqrt{3}}{6}
=-\frac{4\sqrt{3}}{6}
=-\frac{2\sqrt{3}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3}}{2*3}=\frac{0+4\sqrt{3}}{6}
=\frac{4\sqrt{3}}{6}
=\frac{2\sqrt{3}}{3}
$