2(w-3)5w=3(w+3)

Solution for 2(w-3)5w=3(w+3) equation:

2(w-3)5w=3(w+3)

We move all terms to the left:

2(w-3)5w-(3(w+3))=0

We multiply parentheses

10w^2-30w-(3(w+3))=0


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

3(w+3)

We multiply parentheses

3w+9

Back to the equation:

-(3w+9)


We get rid of parentheses

10w^2-30w-3w-9=0

We add all the numbers together, and all the variables

10w^2-33w-9=0

a = 10; b = -33; c = -9;
Δ = b2-4ac
Δ = -332-4·10·(-9)
Δ = 1449
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:

$\sqrt{\Delta}=\sqrt{1449}=\sqrt{9*161}=\sqrt{9}*\sqrt{161}=3\sqrt{161}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-33)-3\sqrt{161}}{2*10}=\frac{33-3\sqrt{161}}{20}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+3\sqrt{161}}{2*10}=\frac{33+3\sqrt{161}}{20}
$