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

Solution for (2w+3)(w-5)=0 equation:

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

We multiply parentheses ..

(+2w^2-10w+3w-15)=0

We get rid of parentheses

2w^2-10w+3w-15=0

We add all the numbers together, and all the variables

2w^2-7w-15=0

a = 2; b = -7; c = -15;
Δ = b2-4ac
Δ = -72-4·2·(-15)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-13}{2*2}=\frac{-6}{4}
=-1+1/2
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+13}{2*2}=\frac{20}{4}
=5
$