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

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

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

We move all terms to the left:

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

Wy multiply elements

4x-((2x-8)*x-1)=0


We calculate terms in parentheses: -((2x-8)*x-1), so:

(2x-8)*x-1

We multiply parentheses

2x^2-8x-1

Back to the equation:

-(2x^2-8x-1)


We get rid of parentheses

-2x^2+4x+8x+1=0

We add all the numbers together, and all the variables

-2x^2+12x+1=0

a = -2; b = 12; c = +1;
Δ = b2-4ac
Δ = 122-4·(-2)·1
Δ = 152
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:

$\sqrt{\Delta}=\sqrt{152}=\sqrt{4*38}=\sqrt{4}*\sqrt{38}=2\sqrt{38}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{38}}{2*-2}=\frac{-12-2\sqrt{38}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{38}}{2*-2}=\frac{-12+2\sqrt{38}}{-4}
$