7w-40=(4w-18)+(w+19)(1/2)

Solution for 7w-40=(4w-18)+(w+19)(1/2) equation:

7w-40=(4w-18)+(w+19)(1/2)

We move all terms to the left:

7w-40-((4w-18)+(w+19)(1/2))=0

We add all the numbers together, and all the variables

7w-((4w-18)+(w+19)(+1/2))-40=0

We multiply parentheses ..

-((4w-18)+(+w^2+19*1/2))+7w-40=0

We multiply all the terms by the denominator


-((4w-18)+(+w^2+19*1+7w*2))-40*2))=0


We calculate terms in parentheses: -((4w-18)+(+w^2+19*1+7w*2)), so:

(4w-18)+(+w^2+19*1+7w*2)

determiningTheFunctionDomain
(+w^2+19*1+7w*2)+(4w-18)

We get rid of parentheses

w^2+7w*2+4w-18+19*1

We add all the numbers together, and all the variables

w^2+4w+7w*2+1

Wy multiply elements

w^2+4w+14w+1

We add all the numbers together, and all the variables

w^2+18w+1

Back to the equation:

-(w^2+18w+1)


We add all the numbers together, and all the variables

-(w^2+18w+1)=0

We get rid of parentheses

-w^2-18w-1=0

We add all the numbers together, and all the variables

-1w^2-18w-1=0

a = -1; b = -18; c = -1;
Δ = b2-4ac
Δ = -182-4·(-1)·(-1)
Δ = 320
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:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-8\sqrt{5}}{2*-1}=\frac{18-8\sqrt{5}}{-2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+8\sqrt{5}}{2*-1}=\frac{18+8\sqrt{5}}{-2}
$