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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+10x^2+4x+5x+2)-((2x-2)(+x))=0


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

(2x-2)(+x)

We multiply parentheses ..

(+2x^2-2x)

We get rid of parentheses

2x^2-2x

Back to the equation:

-(2x^2-2x)


We get rid of parentheses

10x^2-2x^2+4x+5x+2x+2=0

We add all the numbers together, and all the variables

8x^2+11x+2=0

a = 8; b = 11; c = +2;
Δ = b2-4ac
Δ = 112-4·8·2
Δ = 57
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{-(11)-\sqrt{57}}{2*8}=\frac{-11-\sqrt{57}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{57}}{2*8}=\frac{-11+\sqrt{57}}{16}
$