(w-9)(w-3)+(2w-7)(w+8)=w(2w-6)+41

Solution for (w-9)(w-3)+(2w-7)(w+8)=w(2w-6)+41 equation:

(w-9)(w-3)+(2w-7)(w+8)=w(2w-6)+41

We move all terms to the left:

(w-9)(w-3)+(2w-7)(w+8)-(w(2w-6)+41)=0

We multiply parentheses ..

(+w^2-3w-9w+27)+(2w-7)(w+8)-(w(2w-6)+41)=0


We calculate terms in parentheses: -(w(2w-6)+41), so:

w(2w-6)+41

We multiply parentheses

2w^2-6w+41

Back to the equation:

-(2w^2-6w+41)


We get rid of parentheses

w^2-2w^2-3w-9w+(2w-7)(w+8)+6w+27-41=0

We multiply parentheses ..

w^2-2w^2+(+2w^2+16w-7w-56)-3w-9w+6w+27-41=0

We add all the numbers together, and all the variables

-1w^2+(+2w^2+16w-7w-56)-6w-14=0

We get rid of parentheses

-1w^2+2w^2+16w-7w-6w-56-14=0

We add all the numbers together, and all the variables

w^2+3w-70=0

a = 1; b = 3; c = -70;
Δ = b2-4ac
Δ = 32-4·1·(-70)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-17}{2*1}=\frac{-20}{2}
=-10
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+17}{2*1}=\frac{14}{2}
=7
$