20=(1/2)(130-x)

Solution for 20=(1/2)(130-x) equation:

20=(1/2)(130-x)

We move all terms to the left:

20-((1/2)(130-x))=0


Domain of the equation: 2)(130-x))!=0

x∈R


We add all the numbers together, and all the variables

-((+1/2)(-1x+130))+20=0

We multiply parentheses ..

-((-1x^2+1/2*130))+20=0

We multiply all the terms by the denominator


-((-1x^2+1+20*2*130))=0


We calculate terms in parentheses: -((-1x^2+1+20*2*130)), so:

(-1x^2+1+20*2*130)

We get rid of parentheses

-1x^2+1+20*2*130

We add all the numbers together, and all the variables

-1x^2+5201

Back to the equation:

-(-1x^2+5201)


We get rid of parentheses

1x^2-5201=0

We add all the numbers together, and all the variables

x^2-5201=0

a = 1; b = 0; c = -5201;
Δ = b2-4ac
Δ = 02-4·1·(-5201)
Δ = 20804
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:

$\sqrt{\Delta}=\sqrt{20804}=\sqrt{4*5201}=\sqrt{4}*\sqrt{5201}=2\sqrt{5201}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{5201}}{2*1}=\frac{0-2\sqrt{5201}}{2}
=-\frac{2\sqrt{5201}}{2}
=-\sqrt{5201}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{5201}}{2*1}=\frac{0+2\sqrt{5201}}{2}
=\frac{2\sqrt{5201}}{2}
=\sqrt{5201}
$