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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+z^2+9z+4z+36)=0

We get rid of parentheses

z^2+9z+4z+36=0

We add all the numbers together, and all the variables

z^2+13z+36=0

a = 1; b = 13; c = +36;
Δ = b2-4ac
Δ = 132-4·1·36
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-5}{2*1}=\frac{-18}{2}
=-9
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+5}{2*1}=\frac{-8}{2}
=-4
$