-1/2x+13=x-5

Solution for -1/2x+13=x-5 equation:

-1/2x+13=x-5

We move all terms to the left:

-1/2x+13-(x-5)=0


Domain of the equation: 2x!=0

x!=0/2

x!=0

x∈R


We get rid of parentheses

-1/2x-x+5+13=0

We multiply all the terms by the denominator


-x*2x+5*2x+13*2x-1=0

Wy multiply elements

-2x^2+10x+26x-1=0

We add all the numbers together, and all the variables

-2x^2+36x-1=0

a = -2; b = 36; c = -1;
Δ = b2-4ac
Δ = 362-4·(-2)·(-1)
Δ = 1288
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:

$\sqrt{\Delta}=\sqrt{1288}=\sqrt{4*322}=\sqrt{4}*\sqrt{322}=2\sqrt{322}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-2\sqrt{322}}{2*-2}=\frac{-36-2\sqrt{322}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+2\sqrt{322}}{2*-2}=\frac{-36+2\sqrt{322}}{-4}
$