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

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

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

We move all terms to the left:

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

We multiply parentheses

2x^2+8x-((x-5)(x-5))=0

We multiply parentheses ..

2x^2-((+x^2-5x-5x+25))+8x=0


We calculate terms in parentheses: -((+x^2-5x-5x+25)), so:

(+x^2-5x-5x+25)

We get rid of parentheses

x^2-5x-5x+25

We add all the numbers together, and all the variables

x^2-10x+25

Back to the equation:

-(x^2-10x+25)


We add all the numbers together, and all the variables

2x^2+8x-(x^2-10x+25)=0

We get rid of parentheses

2x^2-x^2+8x+10x-25=0

We add all the numbers together, and all the variables

x^2+18x-25=0

a = 1; b = 18; c = -25;
Δ = b2-4ac
Δ = 182-4·1·(-25)
Δ = 424
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{424}=\sqrt{4*106}=\sqrt{4}*\sqrt{106}=2\sqrt{106}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{106}}{2*1}=\frac{-18-2\sqrt{106}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{106}}{2*1}=\frac{-18+2\sqrt{106}}{2}
$