2(w+1)4w=3(2w-1)+8

Solution for 2(w+1)4w=3(2w-1)+8 equation:

2(w+1)4w=3(2w-1)+8

We move all terms to the left:

2(w+1)4w-(3(2w-1)+8)=0

We multiply parentheses

8w^2+8w-(3(2w-1)+8)=0


We calculate terms in parentheses: -(3(2w-1)+8), so:

3(2w-1)+8

We multiply parentheses

6w-3+8

We add all the numbers together, and all the variables

6w+5

Back to the equation:

-(6w+5)


We get rid of parentheses

8w^2+8w-6w-5=0

We add all the numbers together, and all the variables

8w^2+2w-5=0

a = 8; b = 2; c = -5;
Δ = b2-4ac
Δ = 22-4·8·(-5)
Δ = 164
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:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{41}}{2*8}=\frac{-2-2\sqrt{41}}{16}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{41}}{2*8}=\frac{-2+2\sqrt{41}}{16}
$