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

Solution for 2(w+2-)w=3(w-1)+9 equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

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


We calculate terms in parentheses: -(3(w-1)+9), so:

3(w-1)+9

We multiply parentheses

3w-3+9

We add all the numbers together, and all the variables

3w+6

Back to the equation:

-(3w+6)


We get rid of parentheses

2w^2-3w-6=0

a = 2; b = -3; c = -6;
Δ = b2-4ac
Δ = -32-4·2·(-6)
Δ = 57
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{-(-3)-\sqrt{57}}{2*2}=\frac{3-\sqrt{57}}{4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{57}}{2*2}=\frac{3+\sqrt{57}}{4}
$