(2z+1)(9+z)=0

Solution for (2z+1)(9+z)=0 equation:

(2z+1)(9+z)=0

We add all the numbers together, and all the variables

(2z+1)(z+9)=0

We multiply parentheses ..

(+2z^2+18z+z+9)=0

We get rid of parentheses

2z^2+18z+z+9=0

We add all the numbers together, and all the variables

2z^2+19z+9=0

a = 2; b = 19; c = +9;
Δ = b2-4ac
Δ = 192-4·2·9
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-17}{2*2}=\frac{-36}{4}
=-9
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+17}{2*2}=\frac{-2}{4}
=-1/2
$