(2z+5)(9+z)=0

Solution for (2z+5)(9+z)=0 equation:

(2z+5)(9+z)=0

We add all the numbers together, and all the variables

(2z+5)(z+9)=0

We multiply parentheses ..

(+2z^2+18z+5z+45)=0

We get rid of parentheses

2z^2+18z+5z+45=0

We add all the numbers together, and all the variables

2z^2+23z+45=0

a = 2; b = 23; c = +45;
Δ = b2-4ac
Δ = 232-4·2·45
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-13}{2*2}=\frac{-36}{4}
=-9
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+13}{2*2}=\frac{-10}{4}
=-2+1/2
$