10-8x=(6x-4)2x

Solution for 10-8x=(6x-4)2x equation:

10-8x=(6x-4)2x

We move all terms to the left:

10-8x-((6x-4)2x)=0


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

(6x-4)2x

We multiply parentheses

12x^2-8x

Back to the equation:

-(12x^2-8x)


We get rid of parentheses

-12x^2-8x+8x+10=0

We add all the numbers together, and all the variables

-12x^2+10=0

a = -12; b = 0; c = +10;
Δ = b2-4ac
Δ = 02-4·(-12)·10
Δ = 480
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:

$\sqrt{\Delta}=\sqrt{480}=\sqrt{16*30}=\sqrt{16}*\sqrt{30}=4\sqrt{30}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{30}}{2*-12}=\frac{0-4\sqrt{30}}{-24}
=-\frac{4\sqrt{30}}{-24}
=-\frac{\sqrt{30}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{30}}{2*-12}=\frac{0+4\sqrt{30}}{-24}
=\frac{4\sqrt{30}}{-24}
=\frac{\sqrt{30}}{-6}
$