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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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


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

5(2x-5)x

We multiply parentheses

10x^2-25x

Back to the equation:

-(10x^2-25x)


We add all the numbers together, and all the variables

x-(10x^2-25x)-1=0

We get rid of parentheses

-10x^2+x+25x-1=0

We add all the numbers together, and all the variables

-10x^2+26x-1=0

a = -10; b = 26; c = -1;
Δ = b2-4ac
Δ = 262-4·(-10)·(-1)
Δ = 636
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:

$\sqrt{\Delta}=\sqrt{636}=\sqrt{4*159}=\sqrt{4}*\sqrt{159}=2\sqrt{159}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-2\sqrt{159}}{2*-10}=\frac{-26-2\sqrt{159}}{-20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+2\sqrt{159}}{2*-10}=\frac{-26+2\sqrt{159}}{-20}
$