(2z+1)(4+z)=0

Solution for (2z+1)(4+z)=0 equation:

(2z+1)(4+z)=0

We add all the numbers together, and all the variables

(2z+1)(z+4)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2z^2+9z+4=0

a = 2; b = 9; c = +4;
Δ = b2-4ac
Δ = 92-4·2·4
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{-(9)-7}{2*2}=\frac{-16}{4}
=-4
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+7}{2*2}=\frac{-2}{4}
=-1/2
$