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

Solution for (2x^2-4x+1)=(5x+x^2-1) equation:

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

We move all terms to the left:

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

We get rid of parentheses

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


We calculate terms in parentheses: -((5x+x^2-1)), so:

(5x+x^2-1)

We get rid of parentheses

x^2+5x-1

Back to the equation:

-(x^2+5x-1)


We add all the numbers together, and all the variables

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

We get rid of parentheses

2x^2-x^2-4x-5x+1+1=0

We add all the numbers together, and all the variables

x^2-9x+2=0

a = 1; b = -9; c = +2;
Δ = b2-4ac
Δ = -92-4·1·2
Δ = 73
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{-(-9)-\sqrt{73}}{2*1}=\frac{9-\sqrt{73}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{73}}{2*1}=\frac{9+\sqrt{73}}{2}
$