2w-5-18w=(-1/2)(22w+10)

Solution for 2w-5-18w=(-1/2)(22w+10) equation:

2w-5-18w=(-1/2)(22w+10)

We move all terms to the left:

2w-5-18w-((-1/2)(22w+10))=0


Domain of the equation: 2)(22w+10))!=0

w∈R


We add all the numbers together, and all the variables

-16w-((-1/2)(22w+10))-5=0

We multiply parentheses ..

-((-22w^2-1/2*10))-16w-5=0

We multiply all the terms by the denominator


-((-22w^2-1-16w*2*10))-5*2*10))=0


We calculate terms in parentheses: -((-22w^2-1-16w*2*10)), so:

(-22w^2-1-16w*2*10)

We get rid of parentheses

-22w^2-16w*2*10-1

Wy multiply elements

-22w^2-320w*1-1

Wy multiply elements

-22w^2-320w-1

Back to the equation:

-(-22w^2-320w-1)


We add all the numbers together, and all the variables

-(-22w^2-320w-1)=0

We get rid of parentheses

22w^2+320w+1=0

a = 22; b = 320; c = +1;
Δ = b2-4ac
Δ = 3202-4·22·1
Δ = 102312
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{102312}=\sqrt{1764*58}=\sqrt{1764}*\sqrt{58}=42\sqrt{58}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(320)-42\sqrt{58}}{2*22}=\frac{-320-42\sqrt{58}}{44}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(320)+42\sqrt{58}}{2*22}=\frac{-320+42\sqrt{58}}{44}
$