w+(1/2w+6)-(432)=0

Solution for w+(1/2w+6)-(432)=0 equation:

w+(1/2w+6)-(432)=0


Domain of the equation: 2w+6)!=0

w∈R


We get rid of parentheses

w+1/2w+6-432=0

We multiply all the terms by the denominator


w*2w+6*2w-432*2w+1=0

Wy multiply elements

2w^2+12w-864w+1=0

We add all the numbers together, and all the variables

2w^2-852w+1=0

a = 2; b = -852; c = +1;
Δ = b2-4ac
Δ = -8522-4·2·1
Δ = 725896
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:

$\sqrt{\Delta}=\sqrt{725896}=\sqrt{4*181474}=\sqrt{4}*\sqrt{181474}=2\sqrt{181474}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-852)-2\sqrt{181474}}{2*2}=\frac{852-2\sqrt{181474}}{4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-852)+2\sqrt{181474}}{2*2}=\frac{852+2\sqrt{181474}}{4}
$