3w(7w+12)=2(w-3)

Solution for 3w(7w+12)=2(w-3) equation:

3w(7w+12)=2(w-3)

We move all terms to the left:

3w(7w+12)-(2(w-3))=0

We multiply parentheses

21w^2+36w-(2(w-3))=0


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

2(w-3)

We multiply parentheses

2w-6

Back to the equation:

-(2w-6)


We get rid of parentheses

21w^2+36w-2w+6=0

We add all the numbers together, and all the variables

21w^2+34w+6=0

a = 21; b = 34; c = +6;
Δ = b2-4ac
Δ = 342-4·21·6
Δ = 652
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:

$\sqrt{\Delta}=\sqrt{652}=\sqrt{4*163}=\sqrt{4}*\sqrt{163}=2\sqrt{163}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-2\sqrt{163}}{2*21}=\frac{-34-2\sqrt{163}}{42}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+2\sqrt{163}}{2*21}=\frac{-34+2\sqrt{163}}{42}
$