18-(x+3)-2x=4+(x-5)x=

Solution for 18-(x+3)-2x=4+(x-5)x= equation:

18-(x+3)-2x=4+(x-5)x=

We move all terms to the left:

18-(x+3)-2x-(4+(x-5)x)=0

We add all the numbers together, and all the variables

-2x-(x+3)-(4+(x-5)x)+18=0

We get rid of parentheses

-2x-x-(4+(x-5)x)-3+18=0


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

4+(x-5)x

determiningTheFunctionDomain
(x-5)x+4

We multiply parentheses

x^2-5x+4

Back to the equation:

-(x^2-5x+4)


We add all the numbers together, and all the variables

-3x-(x^2-5x+4)+15=0

We get rid of parentheses

-x^2-3x+5x-4+15=0

We add all the numbers together, and all the variables

-1x^2+2x+11=0

a = -1; b = 2; c = +11;
Δ = b2-4ac
Δ = 22-4·(-1)·11
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4\sqrt{3}}{2*-1}=\frac{-2-4\sqrt{3}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4\sqrt{3}}{2*-1}=\frac{-2+4\sqrt{3}}{-2}
$