x=(2x-1)(8x-5)

Solution for x=(2x-1)(8x-5) equation:

x=(2x-1)(8x-5)

We move all terms to the left:

x-((2x-1)(8x-5))=0

We multiply parentheses ..

-((+16x^2-10x-8x+5))+x=0


We calculate terms in parentheses: -((+16x^2-10x-8x+5)), so:

(+16x^2-10x-8x+5)

We get rid of parentheses

16x^2-10x-8x+5

We add all the numbers together, and all the variables

16x^2-18x+5

Back to the equation:

-(16x^2-18x+5)


We add all the numbers together, and all the variables

x-(16x^2-18x+5)=0

We get rid of parentheses

-16x^2+x+18x-5=0

We add all the numbers together, and all the variables

-16x^2+19x-5=0

a = -16; b = 19; c = -5;
Δ = b2-4ac
Δ = 192-4·(-16)·(-5)
Δ = 41
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-\sqrt{41}}{2*-16}=\frac{-19-\sqrt{41}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{41}}{2*-16}=\frac{-19+\sqrt{41}}{-32}
$