(2x+1)=(x2-5x+2)

Solution for (2x+1)=(x2-5x+2) equation:

(2x+1)=(x2-5x+2)

We move all terms to the left:

(2x+1)-((x2-5x+2))=0

We add all the numbers together, and all the variables

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

We get rid of parentheses

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


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

(+x^2-5x+2)

We get rid of parentheses

x^2-5x+2

Back to the equation:

-(x^2-5x+2)


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-1x^2+7x-1=0

a = -1; b = 7; c = -1;
Δ = b2-4ac
Δ = 72-4·(-1)·(-1)
Δ = 45
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:

$\sqrt{\Delta}=\sqrt{45}=\sqrt{9*5}=\sqrt{9}*\sqrt{5}=3\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-3\sqrt{5}}{2*-1}=\frac{-7-3\sqrt{5}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+3\sqrt{5}}{2*-1}=\frac{-7+3\sqrt{5}}{-2}
$