(4z+9)(2+z)=0

Solution for (4z+9)(2+z)=0 equation:

(4z+9)(2+z)=0

We add all the numbers together, and all the variables

(4z+9)(z+2)=0

We multiply parentheses ..

(+4z^2+8z+9z+18)=0

We get rid of parentheses

4z^2+8z+9z+18=0

We add all the numbers together, and all the variables

4z^2+17z+18=0

a = 4; b = 17; c = +18;
Δ = b2-4ac
Δ = 172-4·4·18
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-1}{2*4}=\frac{-18}{8}
=-2+1/4
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+1}{2*4}=\frac{-16}{8}
=-2
$