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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2z^2+11z+9=0

a = 2; b = 11; c = +9;
Δ = b2-4ac
Δ = 112-4·2·9
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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