1x+(8/9x)=12

Solution for 1x+(8/9x)=12 equation:

1x+(8/9x)=12

We move all terms to the left:

1x+(8/9x)-(12)=0


Domain of the equation: 9x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

1x+(+8/9x)-12=0

We add all the numbers together, and all the variables

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

We get rid of parentheses

x+8/9x-12=0

We multiply all the terms by the denominator


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

Wy multiply elements

9x^2-108x+8=0

a = 9; b = -108; c = +8;
Δ = b2-4ac
Δ = -1082-4·9·8
Δ = 11376
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:

$\sqrt{\Delta}=\sqrt{11376}=\sqrt{144*79}=\sqrt{144}*\sqrt{79}=12\sqrt{79}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-108)-12\sqrt{79}}{2*9}=\frac{108-12\sqrt{79}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-108)+12\sqrt{79}}{2*9}=\frac{108+12\sqrt{79}}{18}
$