1-6x=4x(3-2x)

Solution for 1-6x=4x(3-2x) equation:

1-6x=4x(3-2x)

We move all terms to the left:

1-6x-(4x(3-2x))=0

We add all the numbers together, and all the variables

-6x-(4x(-2x+3))+1=0


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

4x(-2x+3)

We multiply parentheses

-8x^2+12x

Back to the equation:

-(-8x^2+12x)


We get rid of parentheses

8x^2-12x-6x+1=0

We add all the numbers together, and all the variables

8x^2-18x+1=0

a = 8; b = -18; c = +1;
Δ = b2-4ac
Δ = -182-4·8·1
Δ = 292
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:

$\sqrt{\Delta}=\sqrt{292}=\sqrt{4*73}=\sqrt{4}*\sqrt{73}=2\sqrt{73}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{73}}{2*8}=\frac{18-2\sqrt{73}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{73}}{2*8}=\frac{18+2\sqrt{73}}{16}
$