(x-2)(2x-5)=x(6-x)-2

Solution for (x-2)(2x-5)=x(6-x)-2 equation:

(x-2)(2x-5)=x(6-x)-2

We move all terms to the left:

(x-2)(2x-5)-(x(6-x)-2)=0

We add all the numbers together, and all the variables

(x-2)(2x-5)-(x(-1x+6)-2)=0

We multiply parentheses ..

(+2x^2-5x-4x+10)-(x(-1x+6)-2)=0


We calculate terms in parentheses: -(x(-1x+6)-2), so:

x(-1x+6)-2

We multiply parentheses

-1x^2+6x-2

Back to the equation:

-(-1x^2+6x-2)


We get rid of parentheses

2x^2+1x^2-5x-4x-6x+10+2=0

We add all the numbers together, and all the variables

3x^2-15x+12=0

a = 3; b = -15; c = +12;
Δ = b2-4ac
Δ = -152-4·3·12
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-9}{2*3}=\frac{6}{6}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+9}{2*3}=\frac{24}{6}
=4
$