W(3w+2)-2(w^-8)=w(w+5)-6w+7

Solution for W(3w+2)-2(w^-8)=w(w+5)-6w+7 equation:

(3W+2)-2(W^-8)=W(W+5)-6W+7

We move all terms to the left:

(3W+2)-2(W^-8)-(W(W+5)-6W+7)=0

We multiply parentheses

(3W+2)-2W-(W(W+5)-6W+7)+16=0

We get rid of parentheses

3W-2W-(W(W+5)-6W+7)+2+16=0


We calculate terms in parentheses: -(W(W+5)-6W+7), so:

W(W+5)-6W+7

We add all the numbers together, and all the variables

-6W+W(W+5)+7

We multiply parentheses

W^2-6W+5W+7

We add all the numbers together, and all the variables

W^2-1W+7

Back to the equation:

-(W^2-1W+7)


We add all the numbers together, and all the variables

W-(W^2-1W+7)+18=0

We get rid of parentheses

-W^2+W+1W-7+18=0

We add all the numbers together, and all the variables

-1W^2+2W+11=0

a = -1; b = 2; c = +11;
Δ = b2-4ac
Δ = 22-4·(-1)·11
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$W_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4\sqrt{3}}{2*-1}=\frac{-2-4\sqrt{3}}{-2}
$
$W_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4\sqrt{3}}{2*-1}=\frac{-2+4\sqrt{3}}{-2}
$