w(2w+1)=((2w+1)+w)+5

Solution for w(2w+1)=((2w+1)+w)+5 equation:

w(2w+1)=((2w+1)+w)+5

We move all terms to the left:

w(2w+1)-(((2w+1)+w)+5)=0

We multiply parentheses

2w^2+w-(((2w+1)+w)+5)=0


We calculate terms in parentheses: -(((2w+1)+w)+5), so:

((2w+1)+w)+5


We calculate terms in parentheses: +((2w+1)+w), so:

(2w+1)+w

We add all the numbers together, and all the variables

w+(2w+1)

We get rid of parentheses

w+2w+1

We add all the numbers together, and all the variables

3w+1

Back to the equation:

+(3w+1)


We get rid of parentheses

3w+1+5

We add all the numbers together, and all the variables

3w+6

Back to the equation:

-(3w+6)


We get rid of parentheses

2w^2+w-3w-6=0

We add all the numbers together, and all the variables

2w^2-2w-6=0

a = 2; b = -2; c = -6;
Δ = b2-4ac
Δ = -22-4·2·(-6)
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{13}}{2*2}=\frac{2-2\sqrt{13}}{4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{13}}{2*2}=\frac{2+2\sqrt{13}}{4}
$