2(2x+1)=3(x-4)x

Solution for 2(2x+1)=3(x-4)x equation:

2(2x+1)=3(x-4)x

We move all terms to the left:

2(2x+1)-(3(x-4)x)=0

We multiply parentheses

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


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

3(x-4)x

We multiply parentheses

3x^2-12x

Back to the equation:

-(3x^2-12x)


We get rid of parentheses

-3x^2+4x+12x+2=0

We add all the numbers together, and all the variables

-3x^2+16x+2=0

a = -3; b = 16; c = +2;
Δ = b2-4ac
Δ = 162-4·(-3)·2
Δ = 280
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:

$\sqrt{\Delta}=\sqrt{280}=\sqrt{4*70}=\sqrt{4}*\sqrt{70}=2\sqrt{70}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{70}}{2*-3}=\frac{-16-2\sqrt{70}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{70}}{2*-3}=\frac{-16+2\sqrt{70}}{-6}
$