5+x=(24-x)(19-2x)

Solution for 5+x=(24-x)(19-2x) equation:

5+x=(24-x)(19-2x)

We move all terms to the left:

5+x-((24-x)(19-2x))=0

We add all the numbers together, and all the variables

x-((-1x+24)(-2x+19))+5=0

We multiply parentheses ..

-((+2x^2-19x-48x+456))+x+5=0


We calculate terms in parentheses: -((+2x^2-19x-48x+456)), so:

(+2x^2-19x-48x+456)

We get rid of parentheses

2x^2-19x-48x+456

We add all the numbers together, and all the variables

2x^2-67x+456

Back to the equation:

-(2x^2-67x+456)


We add all the numbers together, and all the variables

x-(2x^2-67x+456)+5=0

We get rid of parentheses

-2x^2+x+67x-456+5=0

We add all the numbers together, and all the variables

-2x^2+68x-451=0

a = -2; b = 68; c = -451;
Δ = b2-4ac
Δ = 682-4·(-2)·(-451)
Δ = 1016
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:

$\sqrt{\Delta}=\sqrt{1016}=\sqrt{4*254}=\sqrt{4}*\sqrt{254}=2\sqrt{254}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(68)-2\sqrt{254}}{2*-2}=\frac{-68-2\sqrt{254}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(68)+2\sqrt{254}}{2*-2}=\frac{-68+2\sqrt{254}}{-4}
$