2x(10x-8)=23

Solution for 2x(10x-8)=23 equation:

2x(10x-8)=23

We move all terms to the left:

2x(10x-8)-(23)=0

We multiply parentheses

20x^2-16x-23=0

a = 20; b = -16; c = -23;
Δ = b2-4ac
Δ = -162-4·20·(-23)
Δ = 2096
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:

$\sqrt{\Delta}=\sqrt{2096}=\sqrt{16*131}=\sqrt{16}*\sqrt{131}=4\sqrt{131}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4\sqrt{131}}{2*20}=\frac{16-4\sqrt{131}}{40}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4\sqrt{131}}{2*20}=\frac{16+4\sqrt{131}}{40}
$