4x+2x(x-5)=3(2x-4)

Solution for 4x+2x(x-5)=3(2x-4) equation:

4x+2x(x-5)=3(2x-4)

We move all terms to the left:

4x+2x(x-5)-(3(2x-4))=0

We multiply parentheses

2x^2+4x-10x-(3(2x-4))=0


We calculate terms in parentheses: -(3(2x-4)), so:

3(2x-4)

We multiply parentheses

6x-12

Back to the equation:

-(6x-12)


We add all the numbers together, and all the variables

2x^2-6x-(6x-12)=0

We get rid of parentheses

2x^2-6x-6x+12=0

We add all the numbers together, and all the variables

2x^2-12x+12=0

a = 2; b = -12; c = +12;
Δ = b2-4ac
Δ = -122-4·2·12
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{-(-12)-4\sqrt{3}}{2*2}=\frac{12-4\sqrt{3}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{3}}{2*2}=\frac{12+4\sqrt{3}}{4}
$