8x^2-19x=2x^-x

Solution for 8x^2-19x=2x^-x equation:

8x^2-19x=2x^-x

We move all terms to the left:

8x^2-19x-(2x^-x)=0

We get rid of parentheses

8x^2-19x-2x^+x=0

We add all the numbers together, and all the variables

8x^2-20x=0

a = 8; b = -20; c = 0;
Δ = b2-4ac
Δ = -202-4·8·0
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-20}{2*8}=\frac{0}{16}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+20}{2*8}=\frac{40}{16}
=2+1/2
$