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

Solution for (w-10)(w-3)=0 equation:

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

We multiply parentheses ..

(+w^2-3w-10w+30)=0

We get rid of parentheses

w^2-3w-10w+30=0

We add all the numbers together, and all the variables

w^2-13w+30=0

a = 1; b = -13; c = +30;
Δ = b2-4ac
Δ = -132-4·1·30
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-7}{2*1}=\frac{6}{2}
=3
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+7}{2*1}=\frac{20}{2}
=10
$