1/(8x-20)-16=10x+51

Solution for 1/(8x-20)-16=10x+51 equation:

1/(8x-20)-16=10x+51

We move all terms to the left:

1/(8x-20)-16-(10x+51)=0


Domain of the equation: (8x-20)!=0

We move all terms containing x to the left, all other terms to the right

8x!=20

x!=20/8

x!=2+1/2

x∈R


We get rid of parentheses

1/(8x-20)-10x-51-16=0

We multiply all the terms by the denominator


-10x*(8x-20)-51*(8x-20)-16*(8x-20)+1=0

We multiply parentheses

-80x^2+200x-408x-128x+1020+320+1=0

We add all the numbers together, and all the variables

-80x^2-336x+1341=0

a = -80; b = -336; c = +1341;
Δ = b2-4ac
Δ = -3362-4·(-80)·1341
Δ = 542016
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:

$\sqrt{\Delta}=\sqrt{542016}=\sqrt{576*941}=\sqrt{576}*\sqrt{941}=24\sqrt{941}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-336)-24\sqrt{941}}{2*-80}=\frac{336-24\sqrt{941}}{-160}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-336)+24\sqrt{941}}{2*-80}=\frac{336+24\sqrt{941}}{-160}
$