(2x)=(x-5)(x-23)

Solution for (2x)=(x-5)(x-23) equation:

(2x)=(x-5)(x-23)

We move all terms to the left:

(2x)-((x-5)(x-23))=0

We multiply parentheses ..

-((+x^2-23x-5x+115))+2x=0


We calculate terms in parentheses: -((+x^2-23x-5x+115)), so:

(+x^2-23x-5x+115)

We get rid of parentheses

x^2-23x-5x+115

We add all the numbers together, and all the variables

x^2-28x+115

Back to the equation:

-(x^2-28x+115)


We add all the numbers together, and all the variables

2x-(x^2-28x+115)=0

We get rid of parentheses

-x^2+2x+28x-115=0

We add all the numbers together, and all the variables

-1x^2+30x-115=0

a = -1; b = 30; c = -115;
Δ = b2-4ac
Δ = 302-4·(-1)·(-115)
Δ = 440
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:

$\sqrt{\Delta}=\sqrt{440}=\sqrt{4*110}=\sqrt{4}*\sqrt{110}=2\sqrt{110}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-2\sqrt{110}}{2*-1}=\frac{-30-2\sqrt{110}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+2\sqrt{110}}{2*-1}=\frac{-30+2\sqrt{110}}{-2}
$