2x(x+5)=(x-4)(x-5)

Solution for 2x(x+5)=(x-4)(x-5) equation:

2x(x+5)=(x-4)(x-5)

We move all terms to the left:

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

We multiply parentheses

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

We multiply parentheses ..

2x^2-((+x^2-5x-4x+20))+10x=0


We calculate terms in parentheses: -((+x^2-5x-4x+20)), so:

(+x^2-5x-4x+20)

We get rid of parentheses

x^2-5x-4x+20

We add all the numbers together, and all the variables

x^2-9x+20

Back to the equation:

-(x^2-9x+20)


We add all the numbers together, and all the variables

2x^2+10x-(x^2-9x+20)=0

We get rid of parentheses

2x^2-x^2+10x+9x-20=0

We add all the numbers together, and all the variables

x^2+19x-20=0

a = 1; b = 19; c = -20;
Δ = b2-4ac
Δ = 192-4·1·(-20)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-21}{2*1}=\frac{-40}{2}
=-20
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+21}{2*1}=\frac{2}{2}
=1
$