(2s+9)(s+9)=0

Solution for (2s+9)(s+9)=0 equation:

(2s+9)(s+9)=0

We multiply parentheses ..

(+2s^2+18s+9s+81)=0

We get rid of parentheses

2s^2+18s+9s+81=0

We add all the numbers together, and all the variables

2s^2+27s+81=0

a = 2; b = 27; c = +81;
Δ = b2-4ac
Δ = 272-4·2·81
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$s_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-9}{2*2}=\frac{-36}{4}
=-9
$
$s_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+9}{2*2}=\frac{-18}{4}
=-4+1/2
$