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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+2z^2+9z+8z+36)=0

We get rid of parentheses

2z^2+9z+8z+36=0

We add all the numbers together, and all the variables

2z^2+17z+36=0

a = 2; b = 17; c = +36;
Δ = b2-4ac
Δ = 172-4·2·36
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*2}=\frac{-18}{4}
=-4+1/2
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+1}{2*2}=\frac{-16}{4}
=-4
$