(d-9)(d-2)=0

Solution for (d-9)(d-2)=0 equation:

(d-9)(d-2)=0

We multiply parentheses ..

(+d^2-2d-9d+18)=0

We get rid of parentheses

d^2-2d-9d+18=0

We add all the numbers together, and all the variables

d^2-11d+18=0

a = 1; b = -11; c = +18;
Δ = b2-4ac
Δ = -112-4·1·18
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-7}{2*1}=\frac{4}{2}
=2
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+7}{2*1}=\frac{18}{2}
=9
$