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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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


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

(x-1)(x-2)

We multiply parentheses ..

(+x^2-2x-1x+2)

We get rid of parentheses

x^2-2x-1x+2

We add all the numbers together, and all the variables

x^2-3x+2

Back to the equation:

-(x^2-3x+2)


We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2-6x-7=0

a = 1; b = -6; c = -7;
Δ = b2-4ac
Δ = -62-4·1·(-7)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-8}{2*1}=\frac{-2}{2}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+8}{2*1}=\frac{14}{2}
=7
$