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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+3z^2+12z-1z-4)=0

We get rid of parentheses

3z^2+12z-1z-4=0

We add all the numbers together, and all the variables

3z^2+11z-4=0

a = 3; b = 11; c = -4;
Δ = b2-4ac
Δ = 112-4·3·(-4)
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{-(11)-13}{2*3}=\frac{-24}{6}
=-4
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+13}{2*3}=\frac{2}{6}
=1/3
$