2x(x-3)=(x+5)(x-2)

Solution for 2x(x-3)=(x+5)(x-2) equation:

2x(x-3)=(x+5)(x-2)

We move all terms to the left:

2x(x-3)-((x+5)(x-2))=0

We multiply parentheses

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

We multiply parentheses ..

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


We calculate terms in parentheses: -((+x^2-2x+5x-10)), so:

(+x^2-2x+5x-10)

We get rid of parentheses

x^2-2x+5x-10

We add all the numbers together, and all the variables

x^2+3x-10

Back to the equation:

-(x^2+3x-10)


We add all the numbers together, and all the variables

2x^2-6x-(x^2+3x-10)=0

We get rid of parentheses

2x^2-x^2-6x-3x+10=0

We add all the numbers together, and all the variables

x^2-9x+10=0

a = 1; b = -9; c = +10;
Δ = b2-4ac
Δ = -92-4·1·10
Δ = 41
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{-(-9)-\sqrt{41}}{2*1}=\frac{9-\sqrt{41}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{41}}{2*1}=\frac{9+\sqrt{41}}{2}
$