(w+8)(w-8)=(10w+3)(w-8)

Solution for (w+8)(w-8)=(10w+3)(w-8) equation:

(w+8)(w-8)=(10w+3)(w-8)

We move all terms to the left:

(w+8)(w-8)-((10w+3)(w-8))=0

We use the square of the difference formula

w^2-((10w+3)(w-8))-64=0

We multiply parentheses ..

w^2-((+10w^2-80w+3w-24))-64=0


We calculate terms in parentheses: -((+10w^2-80w+3w-24)), so:

(+10w^2-80w+3w-24)

We get rid of parentheses

10w^2-80w+3w-24

We add all the numbers together, and all the variables

10w^2-77w-24

Back to the equation:

-(10w^2-77w-24)


We get rid of parentheses

w^2-10w^2+77w+24-64=0

We add all the numbers together, and all the variables

-9w^2+77w-40=0

a = -9; b = 77; c = -40;
Δ = b2-4ac
Δ = 772-4·(-9)·(-40)
Δ = 4489
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4489}=67$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(77)-67}{2*-9}=\frac{-144}{-18}
=+8
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(77)+67}{2*-9}=\frac{-10}{-18}
=5/9
$