(2x-5)*2=(2x-1)(2x+1)-10(2x-1)

Solution for (2x-5)*2=(2x-1)(2x+1)-10(2x-1) equation:

(2x-5)*2=(2x-1)(2x+1)-10(2x-1)

We move all terms to the left:

(2x-5)*2-((2x-1)(2x+1)-10(2x-1))=0

We use the square of the difference formula

4x^2+(2x-5)*2+1=0

We multiply parentheses

4x^2+4x-10+1=0

We add all the numbers together, and all the variables

4x^2+4x-9=0

a = 4; b = 4; c = -9;
Δ = b2-4ac
Δ = 42-4·4·(-9)
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{10}}{2*4}=\frac{-4-4\sqrt{10}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{10}}{2*4}=\frac{-4+4\sqrt{10}}{8}
$