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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+x^2-2x+5x-10)-((2x+1)(x-2))=0


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

(2x+1)(x-2)

We multiply parentheses ..

(+2x^2-4x+x-2)

We get rid of parentheses

2x^2-4x+x-2

We add all the numbers together, and all the variables

2x^2-3x-2

Back to the equation:

-(2x^2-3x-2)


We get rid of parentheses

x^2-2x^2-2x+5x+3x-10+2=0

We add all the numbers together, and all the variables

-1x^2+6x-8=0

a = -1; b = 6; c = -8;
Δ = b2-4ac
Δ = 62-4·(-1)·(-8)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2}{2*-1}=\frac{-8}{-2}
=+4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2}{2*-1}=\frac{-4}{-2}
=+2
$