(10x)(4x^-7)=40x^-8

Solution for (10x)(4x^-7)=40x^-8 equation:

(10x)(4x^-7)=40x^-8

We move all terms to the left:

(10x)(4x^-7)-(40x^-8)=0

We multiply parentheses

40x^2-70x-(40x^-8)=0

We get rid of parentheses

40x^2-70x-40x^+8=0

We add all the numbers together, and all the variables

40x^2-110x+8=0

a = 40; b = -110; c = +8;
Δ = b2-4ac
Δ = -1102-4·40·8
Δ = 10820
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:

$\sqrt{\Delta}=\sqrt{10820}=\sqrt{4*2705}=\sqrt{4}*\sqrt{2705}=2\sqrt{2705}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-110)-2\sqrt{2705}}{2*40}=\frac{110-2\sqrt{2705}}{80}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-110)+2\sqrt{2705}}{2*40}=\frac{110+2\sqrt{2705}}{80}
$