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

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

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

We move all terms to the left:

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

We multiply parentheses ..

(+w^2+2w-5w-10)-3=0

We get rid of parentheses

w^2+2w-5w-10-3=0

We add all the numbers together, and all the variables

w^2-3w-13=0

a = 1; b = -3; c = -13;
Δ = b2-4ac
Δ = -32-4·1·(-13)
Δ = 61
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{61}}{2*1}=\frac{3-\sqrt{61}}{2}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{61}}{2*1}=\frac{3+\sqrt{61}}{2}
$