(4w-4)(5w+2)=w-10

Solution for (4w-4)(5w+2)=w-10 equation:

(4w-4)(5w+2)=w-10

We move all terms to the left:

(4w-4)(5w+2)-(w-10)=0

We get rid of parentheses

(4w-4)(5w+2)-w+10=0

We multiply parentheses ..

(+20w^2+8w-20w-8)-w+10=0

We add all the numbers together, and all the variables

(+20w^2+8w-20w-8)-1w+10=0

We get rid of parentheses

20w^2+8w-20w-1w-8+10=0

We add all the numbers together, and all the variables

20w^2-13w+2=0

a = 20; b = -13; c = +2;
Δ = b2-4ac
Δ = -132-4·20·2
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-3}{2*20}=\frac{10}{40}
=1/4
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+3}{2*20}=\frac{16}{40}
=2/5
$