(d-2)(d+3)=0

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

(d-2)(d+3)=0

We multiply parentheses ..

(+d^2+3d-2d-6)=0

We get rid of parentheses

d^2+3d-2d-6=0

We add all the numbers together, and all the variables

d^2+d-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:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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