47/8x+12=9+4x

Solution for 47/8x+12=9+4x equation:

47/8x+12=9+4x

We move all terms to the left:

47/8x+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

47/8x-(4x+9)+12=0

We get rid of parentheses

47/8x-4x-9+12=0

We multiply all the terms by the denominator


-4x*8x-9*8x+12*8x+47=0

Wy multiply elements

-32x^2-72x+96x+47=0

We add all the numbers together, and all the variables

-32x^2+24x+47=0

a = -32; b = 24; c = +47;
Δ = b2-4ac
Δ = 242-4·(-32)·47
Δ = 6592
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:

$\sqrt{\Delta}=\sqrt{6592}=\sqrt{64*103}=\sqrt{64}*\sqrt{103}=8\sqrt{103}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-8\sqrt{103}}{2*-32}=\frac{-24-8\sqrt{103}}{-64}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8\sqrt{103}}{2*-32}=\frac{-24+8\sqrt{103}}{-64}
$