(w+3)(w-4)=-6

Solution for (w+3)(w-4)=-6 equation:

(w+3)(w-4)=-6

We move all terms to the left:

(w+3)(w-4)-(-6)=0

We add all the numbers together, and all the variables

(w+3)(w-4)+6=0

We multiply parentheses ..

(+w^2-4w+3w-12)+6=0

We get rid of parentheses

w^2-4w+3w-12+6=0

We add all the numbers together, and all the variables

w^2-1w-6=0

a = 1; b = -1; c = -6;
Δ = b2-4ac
Δ = -12-4·1·(-6)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-5}{2*1}=\frac{-4}{2}
=-2
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+5}{2*1}=\frac{6}{2}
=3
$