39/8x+12=9+4x

Solution for 39/8x+12=9+4x equation:

39/8x+12=9+4x

We move all terms to the left:

39/8x+12-(9+4x)=0


Domain of the equation: 8x!=0

x!=0/8

x!=0

x∈R


We add all the numbers together, and all the variables

39/8x-(4x+9)+12=0

We get rid of parentheses

39/8x-4x-9+12=0

We multiply all the terms by the denominator


-4x*8x-9*8x+12*8x+39=0

Wy multiply elements

-32x^2-72x+96x+39=0

We add all the numbers together, and all the variables

-32x^2+24x+39=0

a = -32; b = 24; c = +39;
Δ = b2-4ac
Δ = 242-4·(-32)·39
Δ = 5568
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:

$\sqrt{\Delta}=\sqrt{5568}=\sqrt{64*87}=\sqrt{64}*\sqrt{87}=8\sqrt{87}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-8\sqrt{87}}{2*-32}=\frac{-24-8\sqrt{87}}{-64}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8\sqrt{87}}{2*-32}=\frac{-24+8\sqrt{87}}{-64}
$