W(4w+6)+2w=2(2w2+7w-3)

Solution for W(4w+6)+2w=2(2w2+7w-3) equation:

(4W+6)+2W=2(2W^2+7W-3)

We move all terms to the left:

(4W+6)+2W-(2(2W^2+7W-3))=0

We add all the numbers together, and all the variables

2W+(4W+6)-(2(2W^2+7W-3))=0

We get rid of parentheses

2W+4W-(2(2W^2+7W-3))+6=0


We calculate terms in parentheses: -(2(2W^2+7W-3)), so:

2(2W^2+7W-3)

We multiply parentheses

4W^2+14W-6

Back to the equation:

-(4W^2+14W-6)


We add all the numbers together, and all the variables

6W-(4W^2+14W-6)+6=0

We get rid of parentheses

-4W^2+6W-14W+6+6=0

We add all the numbers together, and all the variables

-4W^2-8W+12=0

a = -4; b = -8; c = +12;
Δ = b2-4ac
Δ = -82-4·(-4)·12
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$W_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$W_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$W_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-16}{2*-4}=\frac{-8}{-8}
=1
$
$W_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+16}{2*-4}=\frac{24}{-8}
=-3
$