(4+z)(3z+1)=0

Solution for (4+z)(3z+1)=0 equation:

(4+z)(3z+1)=0

We add all the numbers together, and all the variables

(z+4)(3z+1)=0

We multiply parentheses ..

(+3z^2+z+12z+4)=0

We get rid of parentheses

3z^2+z+12z+4=0

We add all the numbers together, and all the variables

3z^2+13z+4=0

a = 3; b = 13; c = +4;
Δ = b2-4ac
Δ = 132-4·3·4
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-11}{2*3}=\frac{-24}{6}
=-4
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+11}{2*3}=\frac{-2}{6}
=-1/3
$