(3w-5)(w+1)=0

Solution for (3w-5)(w+1)=0 equation:

(3w-5)(w+1)=0

We multiply parentheses ..

(+3w^2+3w-5w-5)=0

We get rid of parentheses

3w^2+3w-5w-5=0

We add all the numbers together, and all the variables

3w^2-2w-5=0

a = 3; b = -2; c = -5;
Δ = b2-4ac
Δ = -22-4·3·(-5)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-8}{2*3}=\frac{-6}{6}
=-1
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+8}{2*3}=\frac{10}{6}
=1+2/3
$