(8x+40)(13x-10)=x

Solution for (8x+40)(13x-10)=x equation:

(8x+40)(13x-10)=x

We move all terms to the left:

(8x+40)(13x-10)-(x)=0

We add all the numbers together, and all the variables

-1x+(8x+40)(13x-10)=0

We multiply parentheses ..

(+104x^2-80x+520x-400)-1x=0

We get rid of parentheses

104x^2-80x+520x-1x-400=0

We add all the numbers together, and all the variables

104x^2+439x-400=0

a = 104; b = 439; c = -400;
Δ = b2-4ac
Δ = 4392-4·104·(-400)
Δ = 359121
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:

$\sqrt{\Delta}=\sqrt{359121}=\sqrt{49*7329}=\sqrt{49}*\sqrt{7329}=7\sqrt{7329}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(439)-7\sqrt{7329}}{2*104}=\frac{-439-7\sqrt{7329}}{208}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(439)+7\sqrt{7329}}{2*104}=\frac{-439+7\sqrt{7329}}{208}
$