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

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

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

We move all terms to the left:

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

We multiply parentheses

3w^2+9w-(2(w-1)+3)=0


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

2(w-1)+3

We multiply parentheses

2w-2+3

We add all the numbers together, and all the variables

2w+1

Back to the equation:

-(2w+1)


We get rid of parentheses

3w^2+9w-2w-1=0

We add all the numbers together, and all the variables

3w^2+7w-1=0

a = 3; b = 7; c = -1;
Δ = b2-4ac
Δ = 72-4·3·(-1)
Δ = 61
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{61}}{2*3}=\frac{-7-\sqrt{61}}{6}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{61}}{2*3}=\frac{-7+\sqrt{61}}{6}
$