4(w+1)w=5(w-1)+9

Solution for 4(w+1)w=5(w-1)+9 equation:

4(w+1)w=5(w-1)+9

We move all terms to the left:

4(w+1)w-(5(w-1)+9)=0

We multiply parentheses

4w^2+4w-(5(w-1)+9)=0


We calculate terms in parentheses: -(5(w-1)+9), so:

5(w-1)+9

We multiply parentheses

5w-5+9

We add all the numbers together, and all the variables

5w+4

Back to the equation:

-(5w+4)


We get rid of parentheses

4w^2+4w-5w-4=0

We add all the numbers together, and all the variables

4w^2-1w-4=0

a = 4; b = -1; c = -4;
Δ = b2-4ac
Δ = -12-4·4·(-4)
Δ = 65
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{-(-1)-\sqrt{65}}{2*4}=\frac{1-\sqrt{65}}{8}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{65}}{2*4}=\frac{1+\sqrt{65}}{8}
$