12(x-4)=10(x-3)2x

Solution for 12(x-4)=10(x-3)2x equation:

12(x-4)=10(x-3)2x

We move all terms to the left:

12(x-4)-(10(x-3)2x)=0

We multiply parentheses

12x-(10(x-3)2x)-48=0


We calculate terms in parentheses: -(10(x-3)2x), so:

10(x-3)2x

We multiply parentheses

20x^2-60x

Back to the equation:

-(20x^2-60x)


We get rid of parentheses

-20x^2+12x+60x-48=0

We add all the numbers together, and all the variables

-20x^2+72x-48=0

a = -20; b = 72; c = -48;
Δ = b2-4ac
Δ = 722-4·(-20)·(-48)
Δ = 1344
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:

$\sqrt{\Delta}=\sqrt{1344}=\sqrt{64*21}=\sqrt{64}*\sqrt{21}=8\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(72)-8\sqrt{21}}{2*-20}=\frac{-72-8\sqrt{21}}{-40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(72)+8\sqrt{21}}{2*-20}=\frac{-72+8\sqrt{21}}{-40}
$